PROBABILITY - WORK SHEET1. A bag contains 2 yellow, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?A. 1/2 B. 10/21 C. 9/11 D. 7/11

Explanation : Total number of balls = 2 + 3 + 2 = 7. Let S be the sample space. n(S) = Total number of ways of drawing 2 balls out of 7 = 7C2

Let E = Event of drawing 2 balls , none of them is blue. n(E) = Number of ways of drawing 2 balls , none of them is blue= Number of ways of drawing 2 balls from the total 5 (=7-2) balls = 5C2

(∵ There are two blue balls in the total 7 balls. Total number of non-blue balls = 7 - 2 = 5)P(E) = n(E)/n(S)=5C2/7C2=(5×4/2×1)/(7×6/2×1)=5×4/7×6=10/21

2. A die is rolled twice. What is the probability of getting a sum equal to 9?A. 2/3 B. 2/9 C. 1/3 D. 1/9Explanation : Total number of outcomes possible when a die is rolled = 6 (∵ any one face out of 6)Hence, total number of outcomes possible when a die is rolled twice, n(S) = 6 × 6 = 36E = Getting a sum of 9 when the two dice fall = {(3, 6), {4, 5}, {5, 4}, (6, 3)}Hence, n(E) = 4, P(E) = n(E)/n(S)=4/36=1/9

3. When tossing two coins once, what is the probability of heads on both the coins?A. 1/4 B. 1/2 C. 3/4 D. 2/3Explanation : Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail)Hence, total number of outcomes possible when two coins are tossed, n(S) = 2 × 2 = 4(∵ Here, S = {HH, HT, TH, TT})E = event of getting heads on both the coins = {HH}, Hence, n(E) = 1, P(E) = n(E)/n(S)=1/4

4. 5 coins are tossed together. What is the probability of getting exactly 2 heads?A. 1/2 B. 5/16 C. 4/11 D. 7/16Explanation : Total number of outcomes possible when a coin is tossed = 2 (∵ Head or Tail)Hence, total number of outcomes possible when 5 coins are tossed, n(S) = 25

E = Event of getting exactly 2 heads when 5 coins are tossedn(E) = Number of ways of getting exactly 2 heads when 5 coins are tossed = 5C2

P(E) = n(E)/n(S)=5C2/25=(5×4/2×1)/25=5×2/25=5/24=5/16

5. John and Dani go for an interview for two vacancies. The probability for the selection of John is 1/3 and whereas the probability for the selection of Dani is 1/5. A. What is the probability that none of them are selected?A. 3/5 B. 7/12 C. 8/15 D. 1/5Explanation : Let A = the event that John is selected and B = the event that Dani is selected.Given that P(A) = 1/3 and P(B) = 1/5

We know that A¯ is the event that A does not occur and B¯ is the event that B does not occurProbability that none of them are selected=P(A¯∩B¯)=P(A¯).P(B¯)=[ 1 - P(A) ][ 1 - P(B)] =(1−1/3)(1−1/5) =2/3×4/5=8/15

B. What is the probability that only one of them is selected?A. 3/5 B. 2/5 C. 1/5 D. 12/15Explanation : Probability that only one of them is selected=P[(A∩B¯)∪(B∩A¯)]=P(A∩B¯)+P(B∩A¯)=P(A)P(B¯)+P(B)P(A¯) =P(A)[1−P(B)]+P(B)[1−P(A)]=1/3(1−1/5)+1/5(1−1/3)=1/3×4/5+1/5×2/3=4/15+2/15=2/5

6. Six persons enter a lift on the ground floor of a nine floor apartment. Assuming that each of them independently and with equal probability can leave the lift at any floor beginning with the first, what is the probability that all the six persons are leaving the lift at different floors?A. 8!/86B. 8!/87 C. 8C6/ 86 D. 8P6/ 86

Explanation : Apart from the ground floor, there are 8 floorsLet's find out the total number of ways in which all the six persons can leave the lift at eight different floorsThe 1st person can leave the lift in any of the 8 floors (8 ways), 2nd person 7 floors (7 ways)3nd person 6 floors (6 ways), ... 6th person 3 floors (3 ways)Total number of ways = 8 × 7 × 6 × 5 × 4 × 3 = 8P6

(In fact, from the definition of permutations itself, we will be able to directly say that the number of ways in which all the six persons can leave the lift at 8 different floors = 8P6)n(E)= Total Number of ways in which all the six persons can leave the lift at eight different floors = 8P6

Now we will find out the total number of ways in which each of the six persons can leave the lift at any of the eight floorsThe 1st person can leave the lift in any of the 8 floors (8 ways), 2nd person 8 floors (8 ways)3nd person 8 floors (8 ways), ..., The 6th person can leave the lift in any of the 8 floors (8 ways)Total number of ways = 8 × 8 × 8 × 8 × 8 × 8 = 86

i.e., The total number of ways in which each the six persons can leave the lift at any of the eight floors = n(S) = 86 , P(E) = n(E)/n(S)=8P6/86

7. Raja has 4 tickets of a lottery for which 12 tickets were sold and 3 prices are to be given. What is the probability that Raja will win at least one price?A. 41/55 B. 43/55 C. 12/55 D. 14/55Explanation : Let E be the event that Raja will not won any pricen(E) = number of ways in which Raja will not won any price = number of ways of taking 4 tickets from 12, none of them win a price = number of ways of taking 4 tickets from 9 (=12-3) = 9C4

n(S)= number of ways of taking 4 tickets from 12 = 12C4

Probability that Raja will not win any price = n(E)/n(S)=9C4/12C4=14/55Probability that Raja will at least one price = 1 - Probability that Raja will not win any price=1−14/55=41/55

8. A basket contains 15 apples and 10 oranges out of which 4 apples and 2 oranges are defective. If a person takes two fruits at random, what is the probability that either both are apples or both are goodA. 221/300 B. 276/300 C. 1/2 D. 312/401Explanation : Number of ways in which two apples can be taken from 15 apples = 15C2

Number of ways in which two fruits can be taken from 25 (=15+10) fruits = 25C2

P(both are apples)= 15C2/25C2

Number of ways in which two fruits taken can be taken from 19 good fruits = 19C2 (∵ Total good fruits = (15+10) – (4+2) = 19)Number of ways in which two fruits can be taken from 25(=15+10) fruits = 25C2

P(Both are good)= 19C2/25C2, P(Both are Apples and both are good) = ?There are 11 (=15-4) good apples. Number of ways in which two apples can be taken from these 11 good apples = 11C2 Number of ways in which two fruits can be taken from 25 (=15+10) fruits = 25C2

P(Both are Apples and both are good) = 11C2/25C2

P(Both are Apples or Both are Good)= P(Both are Apples ) + P(Both are Good) – P(Both are Apples and both are good)15C2/25C2+19C2/25C2−11C2/25C2=15C2+19C2−11C2/25C2 =221/300

9. A boy has several colored balls in his bag - 12 white, 15 orange and 25 pink. The lights are out and it is totally dark. Now, how many balls he must take out to make sure that he has a pair of each color?A. 40 B. 42 C. 6 D. 12Solution: In this problem, the worst case scenario will be as follows.Step 1 : He picks all of same colored balls of the largest group. Largest group is pink. Therefore, he will be picking all 25 pink balls first.Step 2 : Then he picks all of same colored balls of the next largest group. Second largest group is Orange. Therefore, he will be picking 15 orange balls.Step 3 : Now, if he starts picking the remaining balls which are white, he could very well stop at 2 balls as he would be having a pair of white ball anyways.Therefore total picks = 25 + 15 + 2 = 42.Shortcut :If confused, for these kinds of problems you have to add higher values first + 2, leaving the least group. Here least group is white.Therefore our answer will be 25 pink + 15 orange + 2 = 42Hence the answer is 42.

10. A bag contains 12 white and 18 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?A. 18/145 B. 18/29 C. 36/135 D.36/145Solution: The probability that first ball is white: =12C1/30C1=12/30=2/5Since, the ball is not replaced; hence the number of balls left in bag is 29.Hence the probability the second ball is black:=18C1/29C1=18/29Required probability =(2/5)×(18/29) = 36/145

11. A five-digit number is formed by using digits 1,2,3,4 and 5 without repetition. What is the probability that the number is divisible by 4?A. 1/5 B. 5/6 C. 4/5 D. None of theseSolution: A number divisible by 4 formed using the digits 1,2,3,4 and 5 has to have the last two digits 12 or24 or 32 or 52. In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways. A number divisible by 4 can be formed in 6×4=24 ways.Total number that can be formed using the digits 1,2,3,4 and 5 without repetition=5!=120Required probability = 24/120 = 1/5

12. A bag contains 5 red and 3 green balls. Another bag contains 4 red and 6 green balls. If one ball is drawn from each bag. Find the probability that one ball is red and one is green.A. 19/20 B. 3/8 C. 3/20 D. 21/40Solution: Let A be the event that ball selected from the first bag is red and ball selected from second bag is green. Let B be the event that ball selected from the first bag is green and ball selected from second bag is red.P(A)=(5/8)×(6/10)=3/8 and P(B)=(3/8)×(4/10)=3/20Hence required probability =P(A)+P(B)=3/8+3/20= 21/40

13. If the chance that a vessel arrives safely at a port is 9/10 then what is the chance that out of 5 vessels expected at least 4 will arrive safely?A. 14*94/105 B. 15*95/104 C. 14*93/105 D. 14*96/105

Solution: The probability that exactly 4 vessels arrive safely is: 5C4×(9/10)4(1/10) The probability that all 5 arrive safely is (9/10)5

The probability that at-least 4 vessels arrive safely: 5C4×(9/10)4(1/10)+(9/10)5=14×94/105

14. There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?A. 6/7 B. 1/8 C. 3/8 D. 5/9Solution: Total cases of checking in the hotels =43 ways.Cases when 3 men are checking in different hotels =4×3×2=24 ways.Required probability: =24/43 = 3/8

15. Out of 17 applicants 8 boys and 9 girls. Two persons are to be selected for the job. Find the probability that at least one of the selected persons will be a girl.A. 27/34 B. 25/34 C. 19/34 D. 21/34Solution: The events of selection of two person is redefined as (i) first is a girl AND second is a boy OR (ii) first is boy AND second is a girl OR (iii) first is a girl and second is a girl.So the required probability: =(9/17×8/16)+(8/17×9/16)+(9/17×8/16) =9/34+9/34+9/34 =25/34

16. The probability that a arrow fired from a point will hit the target is 1/4. Three such arrows are fired simultaneously towards the target from that very point. What is the probability that the target will be hit?A. 19/64 B. 23/64 C. 27/64 D. 37/64Solution: Probability of a arrow not hitting the target =3/4Probability that none of the 3 arrow will hit the target: =(3/4)3=27/64Probability that the target will hit at least once:=1−27/64 = 37/64

17. Ten tickets are numbered 1,2,3……,10. Six tickets are selected at random one at a time with replacement. The probability that the largest number appearing on the selected ticket is 7 is.A. 76-1/106 B. 66/106 C. 76-66/106 D. 76+66/106

18. If the integers m and n are chosen at random from 1 to 100, then what is the probability that a number of the form 7^m + 7^n is divisible by 5? A. ½ B. ¼ C. 1/6 D. 1/8Solution: 1,3,7,9 can be the last digits of the number with equal probability One can choose 2 numbers from this with replacement = 4*4 ways =16 ways out of which if you select pairs (7,3) (3,7) (1,9) (9,1) the resulting number would be divisible by 5 Probability of that happening = 4/16 =1/4 

19. There are 5 pair of shoes in a cupboard from which 4 shoes are picked at random. Find the probability that there is at least one pair in the pick.A. 8/21 B. 13/21 C. 1/12 D. None of theseSol: Without restriction, the no of ways of choosing 4 shoes out of 5 pairs (i.e. 10 shoes)= n(S)=10C4

The no. of ways in which no pair is selected= n(E)=10*8*6*4So, No pair of shoe= n(S)/ n(E)= 8/21, At least one pair of shoe= 1-8/21= 13/21.

20. A classroom has 3 electric lamps. From a collocation of 10 electric bulbs of which 6 are good, 3 are selected at random and put in the lamps. Find the probability that(i) the room is lighted. (ii) all lamps are burning.A. 1/30 B. 29/30 C. 1/6 D. 5/6Sol: Out of 10 electric bulbs, 3 are selected in 10C3 ways. So, number of possible outcomes= n(S)= 10C3=120

21. The odds in favour of an event A are 3:4. The odds against another independent event B are 7:4. What is the probability that at least one of the events will happen?A. 3/7 B. 4/11 C. 7/11 D. 4/7

22. A and B play a series of 5 games in badminton. A’s chance of winning a single game against B is 2/3. Find the chance that A wins. (i) exactly 3 games (ii) at least 3 gamesA. 80/243 B. 163/243 C. 64/81 D. 17/81

23. Find the probability that a leap year chosen at random will have 53 Sundays.A. 1/7 B. 2/7 C. 1/14 D. 1/49

24. From a bag containing 8 green and 5 red balls, there are drawn one after the other. Find the probability of all three balls being green ifi. the balls drawn are replaced before the next ball is picked. (8/13)3

ii. the balls drawn are not replaced. 8/13×7/12×6/11

25. Three coins are tossed. What is the probability of gettingi.) 2 Tails and 1 Head ii.) 1 Tails and 2 Head iii.) Neither 3 Tails nor 3 Head iv.) Three Heads v.) At least one tail A. 1/4 B. 3/8 C. 2/3 D. 1/8 E. 7/8 F. 3/4