When a certain basketball player takes his first shot in a game he succeeds with prob. 1/3. If he misses his first shot, he loses confidence and his second shot will go with prob. 1/4. If he misses his first 2 shots, his third shot will go with pro. 1/5. If he misses his first 3 shots, his fourth shot will go with prob. 1/6. If he misses his first 4 shots, the coach will remove hom from the game. Assume that the player keeps shooting until he succeeds or he is removed from the game. Let X denote the number of shots he misses until his first success or until he is removed from the game. Required:
a. Compute the p. m.f. of X
b. Compute E[X]

X : 0 1 2 3 4

P(x) : _ 0.333_0.167 _0.100_0.067__0.333

E(X) = 1.9

Step-by-step explanation:

From the question above:

X ; number of misses

For :

X = 0 ; P(X = 0) = 1/3 = 0.333

X = 1 ; misses first ; succeeds on second

P(X = 1) = 2/3 * 1/4 = 2/12 = 1/6 = 0.167

X = 2 ; misses first, misses second, succeeds on third

P(X = 2) = 2/3 * (1 - 1/4) * 1/5 = 2/3*3/4*1/5 = 6/60 = 1/10 = 0.10

X = 3 :misses first, misses second, misses third succeeds on fourth

P(X = 3) = 2/3 * 3/4 * (1 - 1/5) * 1/6 = 2/3*3/4*4/5*1/6 = 24/360 = 0.067

X= 4

P(X =4) = 2/3 * 3/4 * 4/5 * 5/6 = 120/ 360 = 1/3 = 0.333

X : 0 1 2 3 4

P(x) : _ 0.333_0.167 _0.100_0.067__0.333

E(X) :

Σx*p(x) = (0*0.333) + (1*0.167) + (2*0.100) + (3*0.067) + (4*0.333)

0 + 0.167 + 0.2 + 0.201 + 1.332

= 1.9

(23*2)-12=x     x=34

step-by-step explanation: