In a certain game, you perform three tasks. you flip a...

swerve wrote:In a certain game, you perform three tasks. You flip a quarter, and success would be heads. You roll a single die, and success would be a six. You pick a card from a full playing-card deck, and success would be picking a spades card. If exactly one of these three tasks is successful, then you win the game. What is the probability of winning?

A. 1/48
B. 5/16
C. 11/12
D. 11/16
E. 23/48

The OA is E.

Please, can any expert explain this PS question for me? I can't get the correct answer. I need your help. Thanks.

This is a very strange game, indeed. So the idea is that you "win" if exactly one of the three contests is a success. Meaning: there are three ways you can win.

1) Coin = success/ die = fail / card = fail -
2) Coin = Fail / Die = success / card = Fail
3) Coin = Fail/Die =fail / card = success

Bearing in mind that a coin flip brings a 1/2 probability of success, the die roll brings a 1/6 probability of success (and a 5/6 probability of failure); and the card pick brings a 1/4 probability of success (and a 3/4 probability of failure,) let's evaluate each in turn

1) Coin = success/ die = fail / card = fail ---> (1/2)(5/6)(3/4) = 15/48
2) Coin = Fail / Die = success / card = Fail ---> (1/2)(1/6)(3/4) = 3/48
3) Coin = Fail/Die =fail / card = success --->(1/2)(5/6)(1/4) = 5/48

Add 'em up (15/48) + (3/48) + (5/48) = 23/48. The answer is E

Probability of winning= probability that only one of the three tasks is successful
(i) flipping a quarter (heads - success)--- (A)
prob. of success (A)=1/2, probability of losing (A)=1/2

(ii) rolling a single die, (six-success)--- (B)
prob. of success (B) = 1/6, prob of losing (B)=5/6

(iii) picking a card from a desk (spades-success)----(C)
prob. of success (C) =13/52 =1/4, probability of losing (C)= 1 - (1/4) = 3/4

$$probability\ of\ winning=A\ B.C.\ +A.\ B\ C.+A.B.\ C$$
$$=\left(\frac{1}{2}\cdot\frac{5}{6}\cdot\frac{3}{4}\right)+\left(\frac{1}{2}\cdot\frac{1}{6}\cdot\frac{3}{4}\right)+\left(\frac{1}{2}\cdot\frac{5}{6}\cdot\frac{1}{4}\right)$$
$$\frac{15}{48}+\frac{3}{48}+\frac{5}{48}=\frac{23}{48}\ \left(Option\ E\right)$$