Chap13&14 Review Problems Solution

Stat 21 Instructor : Shobhana M. Stoyanov Fall 2013

Chap 13 & 14 Review problems solution Chapter 13

1. (a) False (probability is between 0 and 1, « chances » between 0% and 100%) (b) True (the event happens about 9 times out of 10)

2. Option (i) is better: you only have to jump one hurdle and not 2

3. Option (ii): there are more ways to win.

4. The probability is equal to !

!"× !!"× !!"× !!"× !!"≈ !

!,!!!,!!!

5. Yes. If Ticket is white -‐> there are 2 chances out of 3 to get number 1, and 1

chance to get 8

6. (a) True (b) True (see example 2 on page 226) (c) False (the two events are dependent. The probability is 1/52 × 1/51)

7. (a)False (b)False (c) True

(Both sequences are equally likely, having chance !!

! )

8. (a) P(all roll show 3 or more spots) = 𝑃(𝑜𝑛𝑒 𝑑𝑖𝑐𝑒 𝑠ℎ𝑜𝑤𝑠 3 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑠𝑝𝑜𝑡𝑠) !

because each trow is independent of the others. Then, 𝑃 𝑜𝑛𝑒 𝑑𝑖𝑐𝑒 𝑠ℎ𝑜𝑤𝑠 3 𝑜𝑟 𝑚𝑜𝑟𝑒 𝑠𝑝𝑜𝑡𝑠 = 𝑃 𝑑𝑖𝑐𝑒 𝑠ℎ𝑜𝑤𝑠 3, 𝑜𝑟 4, 𝑜𝑟 5, 𝑜𝑟 6 = !

!

(4 possibilities out of 6) Finally, P(all roll show 3 or more spots)= 4/6 ! (b) P(none shows 3 or more)=P(all show 2 or less) = P(all show 1, or 2)= 2/6 ! (similarly as before) (c) P(NOT all rolls show 3 or more)=1-‐P(all rolls show 3 or more)=1-‐(answer in (a)) = 1− 4/6 ! 9. Really similarly than in exercise 8:

(a)P(getting 10 sixes)=P(getting 10 at each attempt)= 1/6 !" (b)P(not getting 10 sixes)=1-‐ P(getting 10 sixes)= 1− 1/6 !" (c) P(all rolls show 5 or less spots) = 5/6 !"


10. Option (ii) is better. You have the same 50-‐50 chance of winning 1 dollar, but you can’t loose.

11. Yes. The way is to define the following tickets: 1 1 ; 1 2 ; 1 2 ; 1 3 ; 3 1 ; 3 2 ; 3 2 ; 3 3

12. The probability is equal to !!""

× !!!× !!"≈ !

!,!!!,!!!

Chapter 14

1. (a) The probability is 𝑃 𝑏𝑜𝑡ℎ 𝑠ℎ𝑜𝑤 3 𝑠𝑝𝑜𝑡𝑠 = 𝑃 𝑓𝑖𝑟𝑠𝑡 𝑜𝑛𝑒 𝑠ℎ𝑜𝑤𝑠 3 ×𝑃 𝑠𝑒𝑐𝑜𝑛𝑑 𝑜𝑛𝑒 𝑠ℎ𝑜𝑤𝑠 3 =!!× !!= 1/36

5we can multiply since they are independent. (b)P(both show the same number)=P(both are 1 OR both are 2 OR both are 3 OR….OR both are 6)=P(both are 1)+P(both are 2)+ …+P(both are 6) = 6× !

!× !!= !

!"

(We can add all the probabilities for different possibilities because the events defining the different possibilities are mutually exclusive (they can’t happen at the same time))

2. From figure 1 in chapter 14, the chance is 2/36

3. (a) False. These events are not mutually exclusive. (b) False. Same thing.

4. Option (i) is better; Even if you miss the first time, you get a second try at the

money.

5. (a) False. This is always false in the general case . They give you here probabilities to verify it: P(A and B)=P(A)×P(B)≠ 0, so they’re not mutually exclusive. The only case where it could be true is the case where one probability of A or of B is equal to 0.

(b) True. This is always true. If they are mutually exclusive, the probability of both happening is 0, so there is a dependence between them, a really strong one. (if one is happening, then the other one happening puts the probability to 0 directly)

6. If you want to find the chance that at least of of two events happens, check if they are mutually exclusive. If they are, you can add the chances. If you want to find the chance that both events will happen, check to see if they are independent; if so, you can multiply the chances.

7. P(2 is drawn at least once)=1− 𝑃 2 𝑖𝑠 𝑛𝑒𝑣𝑒𝑟 𝑑𝑟𝑎𝑤𝑛 = 1− 𝑃 2 𝑛𝑜𝑡 𝑑𝑟𝑎𝑤𝑛 𝑒𝑎𝑐ℎ 𝑡𝑖𝑚𝑒 ! = 1− 3/5 !


(we can multiply because independent, since there is replacement) (It is like Chevalier de Méré, but you really don’t need it to compute the probability)

8. 100%. You can’t avoid it. 9. You can draw a picture like figure 1 to see all the possibilities. Then, you see all

the possibilities verifying each event you re interested in, and add the probabilities, since the different possibilities are mutually exclusive.

(a) the good outcomes are 2 1, 3 1, 3 2. Probability=3/12=25% (b) 3/12=25% (c) 100% -‐(25%+25%)=50%

10. Option (ii) is better. There are 60 rolls, and 60 drawson each roll you have 2

chances in 6 to win 1 dollar, and 4 chances in 6 to get nothing; but on each draw; you have 3 chances in 6 to win 1 dollar and 3 chances in 6 to win nothing.

11. (a) 13/52 × 12/51 × 11/50

(b) 39/52 × 38/51 × 37/50 (c) P(not all diamonds)=1-‐P(all diamonds)=1-‐(answer (a))

12. Both statements are true. Getting ten heads in a row is very unlikely. But If you get nine heads in a row, the chance is 50% to get a tenth head.

13. By trial and error (0.98)!" ≅ 0.503, and (0.98)!" ≅ 0.493. With 34 draws, there is 49.3 % chance of getting a red marble; with 35 draws, the chance is 50.3%. The answer is 35.

14. The chances are the same. Each ticket has the same small chance of winning. If

you win the grand prize with one ticket, you can’t win the grand prize with the other ticket. So, buying the two different tickets doubles the chance, since you can apply the addition rule.

Chap13&14 Review Problems Solution

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