I have seen a very similar question being posted and answered on this website an

Question

I have seen a very similar question being posted and answered on this website and I tried the same method but I am still unable to get the correct answer.

(1 point) A hockey player is to take 3 shots on a certain goalie. The probability he will score a goal on his first shot is 0.35. If he scores on his first shot, the chance he will score on his second shot increases by 0.1; if he misses, the chance that he scores on his second shot decreases by 0.1. This pattern continues to on his third shot: If the player scores on his second shot, the probability he will score on his third shot increases by another 0.1; should he not score on his second shot, the probability of scoring on the third shot decreases by another 0.1 A random variable X counts the number of goals this hockey player scores. (a) Finish the probability distribution of X below. Use four decimals in each of your entries. 2 3 P(X =x) (b) How many goals would you expect this hockey player to score? Enter your answer to four decimals. E(X) = (c) Find the standard deviation the random variable X. Enter your answer to two decimals. SD(X) =

Explanation / Answer

A) P(X = 0) = 0.65 * 0.75 * 0.85 = 0.4144

P(X = 1) = P(score a goal on first shot) + P(score a goal on second shot) + P(score a goal on third short)

= 0.35 * 0.55 * 0.65 + 0.65 * 0.25 * 0.65 + 0.65 * 0.75 * 0.15

= 0.3039

P(X = 2) = P(score goal on first and second shot) + P(score goal on first and third goal) + P(score goal on second and third goal)

= 0.35 * 0.45 * 0.45 + 0.35 * 0.55 * 0.35 + 0.65 * 0.25 * 0.35

= 0.1951

P(X =3) = 0.35*0.45*0.55 = 0.0866

B) E(X) = 0*0.4144 + 1*0.3039 + 2*0.1951 + 3*0.0866 = 0.9539

C) E(X2) = 02*0.4144 + 12*0.3039 + 22*0.1951 + 32*0.0866 = 1.8637

Var(X) = E(X2) - (E(X))2 = 1.8637 - 0.95392 = 0.9538

Sd(X) = sqrt(0.9538) = 0.98

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