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 0.9 and with her left hand is 0.4. 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

The probability that she wins the game if she makes her first attempt with her right hand is 0.396, or 39.6%.

Step-by-step explanation:

To calculate the probability that she wins the game, we need to consider the possible outcomes of her three attempts. Let's assume the player starts with her right hand. She can either hit with her first attempt and hit with her second attempt, or she can miss with her first attempt and hit with her second and third attempts. The probability of the first scenario is 0.9 * 0.4 = 0.36, and the probability of the second scenario is 0.1 * 0.4 * 0.9 = 0.036. Therefore, the probability that she wins the game is 0.36 + 0.036 = 0.396, or 39.6%.

Answer 2

chancing for wining is 54%.

Step-by-step explanation:

possible outcomes can be given as

HHH *

HHM *

HMH

MHH *

HMM

MHM

MMH

MMM

Out of those, only 3 will have the participant win the game:

HHH, HHM, MHH.

Probability (HHH) + Probability (HHM) + Probability (MHH)

since attempts are independant.

Probability (HHH) = 0.9 *0.4 *0.9 = 0.324

Probability (HHM) = 0.9 *0.4 *0.3 = 0.108

Probability (MHH) = 0.3 *0.4 *0.9 = 0.108

Add them up and get = (0.324+ 0.108+0.108) = 0.54

hence, chancing for wining is 54%.