Question

In a game, a participant is given three attempts to hit a ball. On each try, she either scores a hit, H, or a miss, M. The game requires that the player must alternate which hand she uses in successive attempts. That is, if she makes her first attempt with her right hand, she must use her left hand for the second attempt and her right hand for the third. Her chance of scoring a hit with her right hand is .7 and with her left hand is .4. Assume that the results of successive attempts are independent and that she wins the game if she scores at least two hits in a row. If she makes her first attempt with her right hand, what is the probability that she wins the game?

Answer 1

Answer: Probability that she wins the game is 0.364.

Step-by-step explanation:

Since we have given that

There are two possibilities : H for hit, M for miss.

Number of attempts to hit a ball = 3

Probability that she hits with right hand = 0.7

Probability that she hits with left hand = 0.4

Probability that she misses with right hand = 0.3

So, the sample space would be

{HHH, MMM, HHM, MMH, HMH, MHM, HMM, MHH}

So, number of elements in the sample space = 8

We need to find the probability that she wins the game.

According to question,  she wins the game if she scores at least two hits in a row. and  If she makes her first attempt with her right hand,

So, our required probability would be

$P(HHH)+P(HHM)+P(MHH)\\\\=0.7\times 0.4\times 0.7+0.7\times 0.4\times 0.3+0.3\times 0.7\times 0.4\\\\=0.196+0.084+0.084\\\\=0.364$

Hence, probability that she wins the game is 0.364.