Two gun shooters fire at two targets. Each of them fires one shot at his target independently of the other. For the first shooter, probability of hitting the target is 13 and for the second it is 14 . Let X1= number of times the first shooter hits the target X2= number of times the second shooter hits the target Let Z=X1−X2 . Then mean of 2Z+3 is

Since X1,X2∈{0,1},Z=X1−X2∈{−1,0,1} Z=−1 iff X1=0,X2=1⇒P(Z=−1)=16 Z=0 iff X1=0X2=0 or X1=1,X2=1⇒P(Z=0)=712 Z=1 iff X1=1, X2=0⇒P(Z=1)=14

Two gun shooters fire at two targets. Each of them fires one shot at his target independently of the other. For the first shooter, probability of hitting the target is 13 and for the second it is 14 . Let X1= number of times the first shooter hits the target X2= number of times the second shooter hits the target Let Z=X1−X2 . Then mean of 2Z+3 is

Since X1,X2∈{0,1},Z=X1−X2∈{−1,0,1} Z=−1 iff X1=0,X2=1⇒P(Z=−1)=16 Z=0 iff X1=0X2=0 or X1=1,X2=1⇒P(Z=0)=712 Z=1 iff X1=1, X2=0⇒P(Z=1)=14

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Two gun shooters fire at two targets. Each of them fires one shot at his target independently of the other. For the first shooter, probability of hitting the target is $\frac{1}{3}$ and for the second it is $\frac{1}{4}$ . Let $X_{1} =$ number of times the first shooter hits the target $X_{2} =$ number of times the second shooter hits the target Let $Z = X_{1} - X_{2}$ . Then mean of $2 Z + 3$ is

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$\frac{19}{6}$

$\frac{37}{6}$

$\frac{17}{6}$

answer is B.

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Detailed Solution

Since $X_{1}, X_{2} \in {0, 1}, Z = X_{1} - X_{2} \in {- 1, 0, 1}$
$Z = - 1 i f f X_{1} = 0, X_{2} = 1 \Rightarrow P (Z = - 1) = \frac{1}{6} Z = 0 i f f X_{1} = 0 X_{2} = 0 o r X_{1} = 1, X_{2} = 1 \Rightarrow P (Z = 0) = \frac{7}{12} Z = 1 i f f X_{1} = 1, X_{2} = 0 \Rightarrow P (Z = 1) = \frac{1}{4}$