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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]

Respuesta :

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}

[tex]$P(x=0)$[/tex] = probability of the player that does not misses the 1st shot

The probability of the player that he success at the 1st shot = [tex]$\frac{1}{3}$[/tex]

[tex]$P(x=1)$[/tex] = probability of the player that misses the 1st shot but succeeds in his 2nd shot [tex]$=\frac{2}{3} \times \frac{1}{4}$[/tex]

               [tex]$=\frac{1}{6}$[/tex]

[tex]$P(x=2)$[/tex] = probability of the player that misses the 1st two shot but succeeds in the 3rd shot [tex]$=\frac{2}{3} \times \frac{3}{4}\times \frac{1}{5}$[/tex]

                                          [tex]$=\frac{1}{10}$[/tex]

[tex]$P(x=3)$[/tex] = probability of the player that misses the 1st three shot but succeeds in the 4th shot  [tex]$=\frac{2}{3} \times \frac{3}{4}\times \frac{4}{5} \times \frac{1}{6}$[/tex]

                                           [tex]$=\frac{1}{15}$[/tex]

Similarly,

[tex]$P(x)$[/tex] = (probability of missing the 1st four shots) x (probability of removing te player after the 1st four shots missed)

[tex]$=\frac{2}{3} \times \frac{3}{4}\times \frac{4}{5} \times \frac{5}{6} \times 1$[/tex]

[tex]$=\frac{1}{3}$[/tex]

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

X = x            0           1               2               3                 4

P(X=x)        [tex]$\frac{1}{3}$[/tex]             [tex]$\frac{1}{6}$[/tex]               [tex]$\frac{1}{10}$[/tex]             [tex]$\frac{1}{15}$[/tex]                [tex]$\frac{1}{3}$[/tex]          

b). [tex]$E(X) = \sum x \times P(X=x)$[/tex]

              [tex]$= 0\times \frac{1}{3}+1\times \frac{1}{16}+2\times \frac{1}{10}+3\times \frac{1}{15}+4\times \frac{1}{3}$[/tex]

             [tex]$=\frac{114}{60}$[/tex]

            = 1.9

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