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?

Accepted Solution

Answer:The probability that she wins the game is 0.364Step-by-step explanation:Let H = HitLet M = MissP(Hit with right hand) = 0.7P(Hit with right hand) = 1-0.3 = 0.3P(Hit with left hand) = 0.4P(Miss with left hand) = 1-0.4 = 0.6First, we need to highlight possible outcomes.Let SS = Sample Space SS = {HHH, HHM, HMH, MHH, HMM, MHM, MMH, MMM}At this point, we follow the assumption that she starts with her right hand (according to the question)Out of the possible events, only 3 will have the participant win the game:Which are:HHH, HHM and MHH.P (HHH) + P (HHM) + P (MHH)P(HHH) = P(Hit with right) and P(Hit with left) and P(Hit with right)P(HHH) = 0.7 * 0.4 * 0.7P(HHH) = 0.196P(HHM) = P(Hit with right) + P(Hit with left) + P(Miss with right)P(HHM) = 0.7 * 0.4 * 0.3P(HHM) = 0.084P(MHH) = 0.084 (Same as above)Probability that she wins = P (HHH) + P (HHM) + P (MHH) = 0.196 + 0.084 + 0.084Probability that she wins = 0.364..