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]

Solution :

Let the number of the shots that the player while playing misses until the first success = x

The random variable X can take the values {0, 1, 2, 3, 4}

= probability of the player that does not misses the 1st shot

The probability of the player that he success at the 1st shot =

= probability of the player that misses the 1st shot but succeeds in his 2nd shot

= probability of the player that misses the 1st two shot but succeeds in the 3rd shot

= probability of the player that misses the 1st three shot but succeeds in the 4th shot  

Similarly,

= (probability of missing the 1st four shots) x (probability of removing te player after the 1st four shots missed)

a). The p.m.f of the X

X = x            0           1               2               3                 4

P(X=x)                                                                        

b).

            = 1.9

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

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