Chapter 15 Probability

Mathematics For Class XProbability (77)http://www.extramarks.com/class_chapter_ques_browse.php?class=10&subject=Mathematics&chapter=Probability&bro=yes

(Q.1) The probability of getting a number greater than 2 or an even number in a single throw of a fair die is:

( 1 mark )

(a)

(b)

(c) (d) None

View Answer

(Ans) (b)

Explanation :

E {3, 4, 5, 6}

n (E) = 4

S ={1,2,3,4,5,6,}

n (S) = 6

(Q.2) The chance that a non leap year contains 53 Saturday is: ( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (a)

Explanation :

A non leap year contains 365 days is

52 weeks + 1 day

S = {S, M, T, W, Th, F, Sa}

n(S) = 7

E = {Sa}

n(E) =1

(Q.3) In a single throw of two dice, the probability of getting a sum of 10 is: ( 1 mark )

(a)

(b)

(c) (d) NoneView Answer

(Ans) (c)

Explanation :

E = {(4, 6), (5, 5), (6, 4)}

n (E) = 3

n (S) = 6 × 6 = 36

P (E) =

(Q.4) When two dice are thrown, the probability of getting equal numbers is: ( 1 mark )

(a) (b) 1

(c) (d) 0View Answer

(Ans) (a)

Explanation :

E {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

n (E) = 6

n (S) = 6 × 6= 36

(Q.5) The probability for a leap year to have 52 Mondays & 53 Sundays is: ( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (b)

Explanation :

A leap year has 366 days is 52 weeks & 2 days . The 2 days can be chosen in 7 ways. They are

(i) M & Tu (ii) Tu + W (iii)W &Th (iv) Th & F (v) F & Sat (vi) Sat & S (vii) S & M

n (E) = 1 (Sa +S)

n (S) = 7

(Q.6) one card is drawn from a well-shuffld deck of 52 card . The probability that the card drawn ,will be a king of red colour

( 1 mark )

(a) .

(b) .

(c) .

(d) . View Answer

(Ans) (a) .

Explanation :

Total no. of card = 52

No. of king of red colour = 2

Probability of getting a king of red colour = =

(Q.7) From a group of 2 boys and 3 girls, two children are selected at random. Find the probability that both the selected children are girls.

( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (b)

Explanation : Let B1 , B2 be two boys and G1, G2, G3 be three girlsThen, the sample space is given by,S = {B1B2, B1G1, B1G2, B1G3, B2G1, B2G2, B2G3, G1G2, G1G3, G2G3}

Total events = 10Let A = both selected children are girlsA = { G1G2, G1G3, G2G3}Favourable cases = 3

P(A) =

(Q.8) Cards marked with number 13, 14, 15, ………,60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the drawn card is a number which is a perfect square.

( 1 mark )

(a)

(b)

(c) (d) 1/52View Answer

(Ans) (b)

Explanation : Total number of cards = 48“Perfect square” nos. are 16, 25, 36, 49 = 4 cards

P(a perfect square card) =

(Q.9) Two dices are thrown at a time. The probability that the difference of the numbers shown on the dices is 2 is: ( 1 mark )

(a)

(b)


(Ans) (b)

Explanation : E (difference in the numbers is 2) = {(1, 3), (3, 1), (2,4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)}n (E) = 8n (S) = 6 × 6 = 36

(Q.10) A die is thrown. Find the probability of getting a number greater than 4.

( 1 mark )

View Answer

(Ans)

Total number of events = 6

Favourable number of elementary events = 2

Hence the probability = 2/6 = 1/3

(Q.11) A bag contains 5 white and 7 red balls. One ball is drawn at random. What is the probability that the ball drawn is white?

( 1 mark )

View Answer

(Ans)

Total numbers of balls in bag = 5+7 =12

Number of white balls in bag = 5

Probability to draw a white ball = 5/12

(Q.12) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

i. red

( 1 mark )

ii.black ?View Answer

(Ans)

Total numbers of balls in bag = 3+5 =8

Number of red balls in bag = 3

Number of black balls in bag = 5

i. Probability that the balldrawn is a red ball = 3/8

ii. Probability that the balldrawn is a black ball=5/8

(Q.13) Why is tossing a coin considered to be a fair way to decide which team should choose ends in a game of cricket?

( 1 mark )

View Answer

(Ans)

Total number of event = 2

Favourable event for Head = 1

Favourable event for tail = 1

Probability of Head = 1/2

Probability of Tail = 1/2

Probability for both the team to win the toss is same.So it isa fair method to start the game.

(Q.14) Find the probability that a number selected from the number 1 to 25 is not a odd number when each of the given numbers is equally likely to be selected.

( 1 mark )

View Answer

(Ans) Total numbers = 1 to 25 = 25Odd numbers between 1 to 25 1,3,5,7,9,11,13,15,17,19,21,23,25 = 13Probability of a selective number is odd = 13/25

Probability of a selective number is not odd = 1-13/25 = 12/25

(Q.15) 13 defective pens are mixed with 130 good ones. It is not possible to just look at pen and tell whether it is defective or not. One pen is taken out at random from this lot. Determine the probability that the pen taken out is not a defective one?

( 1 mark )

View Answer

(Ans)

Total number of good pens = 130

Defective pens = 13

Total number of pens = 143

Probabilitythata selected pen is defective = 13/143 = 1/11

Probability that a selected pen is not defective = 1-1/11 = 10/11

(Q.16) A child has a die whose six faces show the letters as given below.

The die is thrown once. What is the probability of getting

1. A 2. D

( 1 mark )

View Answer

Ans) Total number of faces = 61. Probability of getting A = 2/6 = 1/3 2. Probability of getting D = 1/6

(Q.17) One card is drawn from a well shuffled deck of 52 cards. Calculate the probability that card will be an ace.

( 1 mark )

(a)

(b)


(Ans) (c) Explanation : Total cards = 52 Total events = 52No. of aces = 4

P (no. of aces) = (Q.18) A child has a block in the shape of a cube with one letter written on each face

as shown below: A B C D E A

The cube is thrown once. What is the probability of getting A?

( 1 mark )

(a)

(b)

(c) (d) None of theseView Answer (Ans) (d) None of these

Explanation : Total number of elementary events = 6There are two faces bearing letter A.

Favourable no. of elementary events = 2Hence, P (Getting A) = 2/6 = 1/3

(Q.19) A book containing 100 pages is opened at random. The probability that a doublet page is found is:

( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (a)

Explanation :

E = {11,22, 33, 44, 55, 66, 77 88, 99}

n (E) = 9

S = {1, 2, 3,…, 100}

n(S) =100

P(E) =

(Q.20) Three letters to each of which corresponds an addressed envelope are placed in the envelops at random. The probability that all letters are placed in the right envelopes is:

( 1 mark )

(a) (b) 0


(Ans) (c)

Explanation :

n (E) = 1

n (S) = 3 × 2 × 1 = 6

P (E)

(Q.21) If a coin is tossed twice, the probability of getting at-least one head is:

( 1 mark )

(a)

(b)

(c) (d) None

View Answer

(Ans) (a)

Explanation :

S = {HH,HT,TH,TT}n(S) = 4E = {HT,TH,HH}n(E) = 3

(Q.22) Two numbers are chosen from 1 to 5. The probability for the two numbers to be consecutive is :

( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (c)

Explanation :

E = 1, 2; 2, 3; 3, 4; 4, 5

n (E) = 4

P(E)

(Q.23) The chance that a non leap year contains 53 Fridays is ( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (d)

Explanation : A leap year always has 52 Friday and 2 days (i) S & M (ii) M & T (iii)Tu &Wed (iv) Wed & Th (v) Th & Fri (vi) Fri & Sat (vii) Sat & SOut of 7 cases, 2 Fridays

P {53Fridays} =

(Q.24) A factory manufacturing car batteries made a survey in the field about the life of these batteries. The data obtained under:

( 1 mark )

If you put a battery of this company in your car, what is the probability that the battery will last for more than 36 months

(a) 0.5(b) 0.74(c) 0.85(d) 0.90View Answer

(Ans) (b) 0.74

Explanation :

Total frequency or the total no. of trials made = 1000

The total number of batteries which last for more than 36 months = 540 + 200 = 740

Now, P (battery will last for more than 36 months)

=

(Q.25) The probability of getting an even number when a die is rolled:

( 1 mark )

(a)

(b)

(c) (d) None

View Answer

(Ans) (b)

Explanation :

E = {2, 4, 6}

n(E) = 3

S = {1, 2, 3, 4, 5, 6}n (S)= 6

P(E)=

(Q.26) The probability for a randomly selected number out of 1, 2, 3, 4, …25 to be a even number is

( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (c) Explanation : E = {2,4,6,8,10,12,16,18,20,22,24}n (E) = 12S = {1,2,3,4…., 25}n (S) = 25

(Q.27) The probability of three coins falling all heads up when tossed simultaneously is: ( 1 mark )

(a)

(b)

(c) (d) None

View Answer

(Ans) (b) Explanation :

E = {HHH}, n (E) = 1S = {HHH, HHT, HTH, HTT,THH,THT,TTH,TTT}n (s) = 8

P(E)

(Q.28) The probability that a vowel selected at random in English language is an ‘o’ is: ( 1 mark )

(a)

(b)


(Ans) (c)

Explanation :

E = {o}

n (E) = 1

S = {a, e, i, o,u}

n (S) = 5

P(E) =

(Q.29) When two dice are thrown, the probability of getting a number always greater than 4 on the second: ( 1 mark )

(a)

(b)


(Ans) (a)

Explanation :

E = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (1,6), (2,6), (3,6), (4,6), (5,6), (6,6)}

n (E) =12

n (S) = 6× 6 =36

P (E) =

(Q.30) In a throw of a dice, the probability of getting a prime no. is: ( 1 mark )

(a)

(b)

(c) (d) 6

View Answer

(Ans) (b)

Explanation :

E = {2, 3, 5}

n(E) = 3

S = {1, 2, 3, 4, 5, 6}

N (S) = 6

(Q.31) From a normal pack of cards, a card is drawn at random. The probability of getting a jack or a king is:

( 1 mark )

(a)

(b)


(Ans) (b)

Explanation :

n (E) = 4 + 4 = 8

n (S) = 52

P(E)

(Q.32) A bag contains 3 white & 5 red balls. If a ball is drawn at random, the probability that the drawn ball to red is:

( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (a)

Explanation :

n (E) = 5

n (S) = 3 + 5 = 8

(Q.33) Two dice are thrown at a time. The probability that the difference of the numbers shown on the dice is 1 is:

( 1 mark )

(a)

(b)

(c) (d) None

View Answer

(Ans) (b)

Explanation :

E = {(1, 2,), (2, 1),(2,3), (3, 2), (3. 4) (4, 3), (4, 5), (5, 4), (5, 6),(6, 5)}

n (E) = 10

n (S) = 6× 6 = 36

(Q.34) The probability for a randomly selected number out of 1, 2, 3, 4, …25 to be a prime number is:

( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (d)

Explanation :

E = {2,3,5,7,11,13,17,19,23}

n (E) = 9

S = {1,2,3,4…., 25}

n (S) = 25

(Q.35) A card is drawn from a pocket of 100 cards numbered 1 to 100. The probability of drawing a number. Which is square is:

( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (b)

Explanation :

E = {1, 4, 9,16,25, 36, 49, 64, 81, 100}

n (E) = 10

S = {1,2,3,….,100}

n (S) =100

(Q.36) The probability that in a family of 3 children there will be atleast one boy is:

( 1 mark )

(a)

(b)

(c)

(d)

View Answer

(Ans) (d)

Explanation :

S ={BBB, BBG, BGB,BGG,GBB,GBG, GGB, GGG}

n(S) = 8

E = {BBB, BBG, BGB, BGG, GBB, GBG, GGB}

n (E) = 7

(Q.37) 400 students of class X of a school appeared in a test of 100 marks in the subject of Math’s & the data about the marks secured is as below:

If the result card of a student be picked up at random, what is the probability that the student has secured more that 50 marks?

( 1 mark )

(a) 0.315(b) 0.325(c) 3.6(d) 30.3View Answer

(Ans) (b) 0.325

Explanation :

Total no. of students is the total frequency = 400

The total no of students who secured more than 50 marks = 100 + 30 =130

Probability that the marks secured are more than 50

(Q.38) Fifty seeds were selected at random from 5 bags A, B, C, D, E of seeds and these were kept under standardized conditions equally favourable to germination. After 20 days , the number of seeds which had germinated in each collection were counted & recorded as follow:

( 1 mark )

What is probability of more than 40 germinated seeds in a bag?(a) 0(b) 0.60(c) 0.5(d) 0.25View Answer

(Ans) (b) 0.60

Explanation :

No. of bags in which more than 40 seeds out of 50 seeds germinated is 3 (These are B, C,E}

Total no. of bags =5

So P (more than 40 germinated seeds in a bag)

(Q.39) A die is thrown 1000 times with the frequencies for the outcomes 1,2,3,4,5,& 6 as given in table belo

What is the probability of getting 1 as out come 9 ?

( 1 mark )

(a) 0.5(b) 0.190(c) 0.180(d) 0.13View Answer

(Ans) (c) 0.180

Explanation :

We denote the event of getting 1 by E1

We have

(Q.40) There are 500 packets in a large box & each packet contains 4 electric devices in it. On testing at the time of packing, it was noted that there are some faulty pieces in the pockets. The data is as below:

( 1 mark )

If one packet is drawn from the box, what is probability that all the four devices in the packet are without any faults?

(a) 0.1(b) 0.2(c) 0.6(d) 0.8View Answer (Ans) (c) 0.6

Explanation : When the packet has all the four device without fault, it means the number of faulty devices in the packet is 0. No. of chances which are favourable to 0 are 300 as given in the table above. Thus the probability of

acket containing all the four devices without any fault =

(Q.41) A bag contains 3 red balls and 5 black balls. A ball is drawm at random from the bag. What is probability that the ball drawn isred ? ( 1 mark )

(a) 1/8(b) 2/8(c) 3/8(d) 5/8 View Answer

(Ans) (c) 3/8

Explanation : Red balls = 3, Black balls = 5 Total balls = 3 + 5 = 8 Total number of events = 8P(a red ball) = 3/8

(Q.42) Three unbiased coins are tossed together. Find the probability of getting exactly two heads up

( 1 mark )

(a)

(b)


(Ans) (b)

Explanation :

E = {HHH}, n (E) = 2E = { HHT, HTH, THH}n (E) = 3

P(E) =

(Q.43) A bag contains 5 white & 7 red balls. If a ball is drawn at random, the probability that the drawn ball to red is

( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (a)

Explanation : n (E) = 7n (S) = 5 + 7 = 12

(Q.44) Out of 400 bulbs in a box, 15 bulbs are defective. One bulb ( 1 mark )

is taken out at random from the box. Find the probability that the drawn bulb is not defective.

(a)

(b)


(Ans) (b)

Explanation : Total number of bulbs in a box = 400Total number of defective bulbs in the box = 15

Total number of non-defective bulbs in the box = 400 – 15 = 385

P (bulb is not defective)=

=

(Q.45) Two numbers are chosen from 1 to 7. The probability for the two numbers to be consecutive is: ( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (d) Explanation : E(pairs of consecutive numbers) = 1, 2; 2, 3; 3, 4; 4, 5; 5, 6; 6, 7n (E) = 6

n (S) =

P(E) (Q.46) The probability of getting an odd number when a die is rolled is:

( 1 mark )

(a)

(b)


(Ans) (b)

Explanation : E (odd numbers) = {1, 3, 5}n(E) = 3S = {1, 2, 3, 4, 5, 6}n (S)= 6

P(E)=

(Q.47) A card is drawn at random from a pack of 52 playing cards. Find the probability that the card is neither an ace nor a king.

( 1 mark )

(a)

(b)


(Ans) (a)

Explanation : Q (neither an ace nor a king)= 1 – P (either an ace or a king)= 1 – [P (an ace) + P (a king)]

= 1 -

= 1 -

(Q.48) From a normal pack of cards, a card is drawn at random. The probability of getting a Spade is:

( 1 mark )

(a)

(b)


(Ans) (c)

Explanation : E = spadesn(E) = 13n(S) = 52

P(E)

(Q.49) A dice is thrown twice. Find the probability of getting doublets.

( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (d)

Explanation : Number of events of getting doubletsE = doublets = {(1,1), (2,2), (3,3), (4,4), (5,5),(6,6)}

n(E) = 6

Hence P (getting a doublet) =

(Q.50) If a letter is chosen at random from the English alphabets, find the probability that the letter is a vowel.

( 1 mark )

(a)

(b)


(Ans) (c)

Explanation : There are 26 letters in English alphabets, having 5 vowels and 21 consonants

Total number of events = 26No. of vowels = 5

P (a vowel) =

(Q.51) A letter is chosen at random from the letters of the word ‘ASSASSINATION’. Find the probability that the letter chosen is a vowel.

( 1 mark )

(a)

(b)

(c) (d) 2/7

View Answer

(Ans) (a) Explanation : Total number of elementary events = 13There are 6 vowels in the word ‘ASSASSINATION’. So, there are 6 ways of selecting a vowel.

Probability of selecting vowel =

(Q.52) Gopi buys a fish from a shop for his aquarium. The ( 1 mark ) shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

(a)

(b)

(c) 1/5(d) 2/7View Answer

(Ans) (b) Explanation : There are 13 = 8 + 5 fish out of which one can be chosen in 13 ways.Total number of elementary events = 13Male fish = 5

Hence, Probability =

(Q.53) Three unbiased coins are tossed together. Find the probability of getting all heads.

( 1 mark )

(a) 0(b) 1

(c)

(d) View Answer

(Ans) (d)

Explanation : Elementary events associated to random experiment of tossing three coins areHHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Total number of elementary events = 8The event “Getting all heads” is said to occur, if the elementary event HHH occurs i.e. HHH is an outcome.

Required probability =

(Q.54) Savita and Hamida are friends. What is the probability that both will have the same birthday?

( 1 mark )

(a) 1

(b)


(Ans) (b)

Explanation : Total number of ways in which Savita and Hamida may have their birthdays = 365 x 365Number of ways in which Savita and Hamida will have same birthday = 365

Probability that Savita and Hamida will have the same birthday =

(Q.55) If the probability of winning a game is 0.3, what is the probability of losing it?

( 1 mark )

(a) 0.4(b) 0.6(c) 0.7(d) 0.9 View Answer

(Ans) (c) 0.7

Explanation : Let E be the event and P(E) be the probability of winning the game and P(Not E) losing it. Then P(E) = 1 – P(Not E)P(E) + P (Not E) = 1P(Not E) = 1 – P(E)P(Not E) = 1 – 0.3P(Not E) = 0.7Hence the probability of losing the game is 0.7.

(Q.56) A bag contains 3 red balls, 5 black balls and 4 white balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is white?

( 1 mark )

(a) 1

(b)

(c)

(d) View Answer

(Ans) (b)

Explanation : Random drawing of balls ensures equally likely outcomes.Total number of outcomes = 3 (red) + 5 (black) + 4 (white) = 12Out of total 12 outcomes, favourable outcome = 4, since the bag contains 4 white balls.

P (a white ball) =

(Q.57) A card is drawn at random from a well shuffled pack of 52 cards. Find the probability that the card is neither a red card nor a queen.

( 1 mark )

(a) 1/15

(b)

(c)

(d) View Answer

(Ans) (b)

Explanation : There are 26 cards, including two red queens in a pack of 52 cards.Also there are 4 queens i.e., two red and two black.Therefore, card drawn will be a red card or a queen if it is any one of28 cards ( 26 red cards and 2 black queens).Now, the card drawn will be neither a red card nor a queen= Total number of cards – The card drawn will be a red or a queen= 52 – 28= 24 cardsSo, the favourable number of elementary events = 24 cards

Hence, the required probability = .

(Q.58) Two dice are thrown simultaneously. Find the probability of getting the sum a multiple of 3 and an even number.

( 1 mark )

(a) 1 / 6(b) 1 / 24(c) 1/ 36(d) 5/ 48View Answer

(Ans) (a) 1 / 6

Explanation :

Number of possible outcomes = 6 x 6 = 36.Number of favourable outcomes = n{(1, 5), (5, 1), (2, 4), (4, 2), (3, 3), (6, 6)} = 6.

Required probability = (Number of favourable outcomes) / (Number of possible outcomes)

= 6 / 36 = 1 / 6.

(Q.59) The probability for a randomly selected number out of 1, 2, 3, ,..…35 to be a multiple of 7 is

( 1 mark )

(a)

(b)

(c)

(d) View Answer

(Ans) (d)

Explanation : E = {7,14,21,28,35}n (E) = 5S = {1,2,3,4…., 25}n (S) = 35

(Q.60) A bag contains 5 Red and 7 Blue balls .The probability of drawing a blue ball from the bag is

( 1 mark )

(a) .

(b) .

(c) .

(d) .View Answer

(Ans) (c) .

Explanation :

No. of Red balls = 5 ,No. of Blue balls = 7Total no. of balls = 5+7 =12

Probability of drawing a Blue ball P(B) =

(Q.61) A coin is tossed 25 times while tail occurred 15 times ,In this situation probabilityof getting a tail is

( 1 mark )

(a) .

(b) .

(c) .

(d) .View Answer

(Ans) (b) .

Explanation :

Probability of getting a tail = number of tails / total number of trials = 15/25 =

(Q.62) The percentage of marks obtained by a students in maths test is as given below -Unit test I II IIIPercentage % 75 80 85Probability that the student secures more than 80% in test is

( 1 mark )

(a) .

(b) .

(c) .

(d) .View Answer

(Ans) (b) .

Explanation :

Total number of test = 3

Number of tests in which the student got more than 80% =1Probability of getting more than 80 % =1/3

(Q.63) The probability of occurring of two heads when two coins are tossed simultaneously

( 1 mark )

(a) .

(b) .

(c) .

(d) .View Answer

(Ans) (c) .

Explanation :

When two coins are tossed simultaneously , As an outcome , HH , HT , TT , TH Favourable no. of events = 1

Probability of occurring of two heads =

(Q.64) The probability of occurring of a multiple of 2 when a dice is thrown

( 1 mark )

(a) .

(b) .

(c) .

(d) . View Answer

(Ans) (d) .

Explanation :

Multiples of 2 are 2 , 4, 6Favourable no. of events =3Total nos. = 6Probability of occurring of a multiple of 2

P ( E ) = 3/6 =

(Q.65) A die is thrown 50 times with the following frequency for the out comes of 1,2,3,4 As- outcomes 1 2 3 4Frequency 10 20 25 30 The probabilities of outcomes less than 3 is

( 1 mark )

(a) 0.2(b) 0.5(c) 0.6(d) 1View Answer

(Ans) (c) 0.6

Explanation :

Total number of throws = 50Total number of occurrence less than 3 = 10 + 20 = 30

Probability of outcomes less than 3 = 30/50 = = 0.6

(Q.66) 7 bags of rice, each marked 10 kg, actually contained the following weights of rice(in Kg.)10.01, 9.97, 10.03, 9.96, 10.04, 10.06, 9.98. Thenthe probability of that any of these bags chosen at random contain less than 10 kg of rice is

( 1 mark )

(a) .

(b) .

(c) .(d) 1.View Answer

(Ans) (a) .

Explanation : Total Numbers of bags = 7Number of bags contain less than 10 kg of Rice = 3

Probability of bags that contain less than 10 kg =

(Q.67) A letter is selected at random from the word “ Elephant “ the probability that the selected letter is a vowel

( 1 mark )

(a) .

(b) .

(c) .

(d) . View Answer

(Ans) (b) .

Explanation : Total no. of letters = 8

No. of vowel in the given word = 3

Probability ( selected is a vowel ) =

(Q.68) A coin is tossed 100 times with the following frequencies: Tail: 55Head: 45, the probability for occurrence of heads is

( 1 mark )

(a) .

(b) .(c) 0.55 .(d) 0.45 .View Answer

Ans) (d) 0.45 .

Explanation :

(Q.69) In a simultaneous throw of a pair of dice, find the probability of getting 8 as sum.

( 3 Marks )

View Answer

(Ans)

Elementary events associated to the random experiment of throwing two dies.

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

Total event = 6x6 = 36Favourable event is when the sum is 8.(2,6),(3,5),(4,4),(5,3),(6,2)Favourable event = 5Hence probability of sum 8 = 5/36

(Q.70) Three coins are tossed together. Find the probability of getting exactly two heads.

( 3 Marks )

View Answer

(Ans)

Elementary events associated to the random experiment of throwing three coins

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT


Favourable events

HHT, HTH, THH

Favourable events = 3

Probability of events = 3/8

(Q.71) There are 30 cards, of same size, in a bag on which numbers 1 to 30 are written. One card is taken out of the bag at random. Find the probability that the number on the selected card is not divisible by 3

( 3 Marks )

View Answer

(Ans)

Total number of event = 30

Favourable condition for selection of a card multiple of 3

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Total condition = 10

Probability to select a card divisible by three P(A) = 10/30 = 1/3

Probability to select a card not divisible by three p(B)

We know that

P(A)+P(B) = 1

P(B) = 1-1/3 = 2/3

(Q.72) A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is (i) red or white (ii) not black (iii) neither white nor black.

( 3 Marks )

View Answer

(Ans)

Total number of balls in bag = 5+8+7 = 20

i. red or white balls

Total number of red and white balls = 5+8 = 13

Probability to drawn red or white = 13/20

ii. Not black balls

Total number of Black balls = 7

Probability to drawn black balls = 7/20

Probability to drawn not a black balls = 1-7/20 = 13/20

iii . Neither white nor black

Total number of balls in bag = 20

White and black = 15

Neither white nor black = 20 – 15 – 5

Probability to drawn neither white nor black = 5/20 = 1/4

(Q.73) A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number,1,2,3,4….16 as shown in figure.What is the probability that it will point to

1. 10

2. An odd number

3. Amultiple of 3

4. an even number.

( 3 Marks )

View Answer

(Ans)


i. Probability to get 10 = 1/16

ii. Total odd numbers = 1,3,5,7,9,11,13,15 = 8

Probability to get an odd number = 8/16 = 1/2

iii Total numbers multiple of 3 = 3,6,9,12,15 =5

Probability to get a number multiple of 3 = 5/16

iv Total even numbers = 2, 4, 6,8,10,12,14,16 = 8

Probability to get a even number = 8/16 = 1/2

(Q.74) A bag contains 8 red, 6 white and 4 black balls. A ball is drawn a random from the bag. Find the probability that the drawn ball is

1. red or white

2. not black

3. neither white nor black.

( 3 Marks )

View Answer

(Ans)

Total number of balls in bag = 8+6+4 = 18

1. red or white balls

Total number of red and white balls = 6+8 = 14

Probability to drawn red or white = 14/18=7/9

2.Not black balls

Total number of Black balls = 4

Probability to drawn black balls = 4/18=2/9

Probability to drawn not a black balls = 1-2/9 = 7/9

3.Neither white nor black

Total number of balls in bag = 18

White and black = 10

Neither white nor black = 18-10 = 8

Probability to drawn neither white nor black = 8/18 = 4/9

(Q.75) Find the probability that a number selected at random from the numbers 1,2,3……,35 is a

1. Multiple of 7

2. Multiple of 3 and 5

3. Multiple of 3 or 5

( 3 Marks )

View Answer

(Ans)

Total numbers = 1 to 35 = 35

i. Number multiple of 7 = 7,14,21,28,35 = 5

Probability to select a number multiple of 7 = 5/35 = 1/7

ii. multiple of 3 = 3,6,9,12,15,18,21,24,27,30,33

Multiple of 5 = 5,10,15,20,25,30,35

Multiple of 3 and 5 = 15,30 = 2

Probability to select a number multiple of 3 and 5 = 2/35

iii. multiple of 3 = 3,6,9,12,15,18,21,24,27,30,33

Multiple of 5 = 5,10,15,20,25,30,35

Multiple of 3 or 5 = 3,5,6,9,10,12,15,18,20,21,24,25,27,30,33,35

Multiple of 3 or 5 = 16

Probability to select a number multiple of 3 or 5 = 16/35

(Q.76) From a pack of 52 cards jacks, queens, kings and aces of red colour are removed. From the remaining, a card is drawn at random. Find the probability that the card drawn is

1. a black queen 2. a red card 3. a black jack 4. A picture card.

( 3 Marks )

View Answer

(Ans)

Total Number of cards = 52

Removed cards = 8

Remaining cards = 44

1. A black queen

Total numbers of queens = 4

Black queens = 2

Probability of black queen = 2/44 = 1/22

2. A red card

Total numbers of red cards = 26

Removed red cards = 8

Remaining red cards = 18

Probability of a red cards = 18/44 = 9/22

3. A black jack

Total numbers of jacks = 4

Black colour jacks = 2

Probability of a black jacks = 2/44 = 1/22

4. A picture card

Total number of picture cards = 12

Removed picture cards = 6

Remaining picture cards = 6

Probability of a picture card = 6/44 = 3/22

(Q.77) A game consists of tossing a one rupee coin 3 times. Hanif wins if all the tosses give the same result that is three heads or three tails, and looses otherwise. Calculate the probability that Hanif will lose the game.

( 3 Marks )

View Answer

(Ans) If a coin is tossed three times total number of possible out comes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Total numbers of outcomes = 8Hanif wins the toss if 3 heads or 3 tails Only possible event for win = 2Probability that Hanif wins = 2/8 =1/4ProbabilitythatHanif does not win = 1-1/4 = 3/4

Chapter 15 Probability

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Transcript of Chapter 15 Probability