Question #142505

A soldier who hits a target with probability 2/3 fires shots until he hits the target or 5 shots have been fired. Let X denote the number of shots fired. Find the
i) probability distribution of X
ii) mean of X

Expert's answer

The number of shots fired


X=1,2,3,4,5X={1,2,3,4,5}

Probability of hitting the target


P=23, (1−P)=13P = \frac{2}{3}, \ (1-P) = \frac{1}{3}

The probability of hitting the target and stopping given X number of shots


P(X=x)=P(1−P)x−1P (X=x) = P(1-P)^{x-1}

i)

P(X=1)=23P (X=1) = \frac{2}{3}P(X=2)=23∗13=29P(X=2) = \frac{2}{3} * \frac{1}{3} = \frac{2}{9}P(X=3)=23∗(13)2=227P(X=3) = \frac{2}{3} * (\frac{1}{3}) ^2 = \frac{2}{27}

P(X=4)=23∗(13)3=281P(X=4) = \frac{2}{3} * (\frac{1}{3}) ^3 = \frac{2}{81}

P(X=5)=23∗(13)4+(13)5=3243P(X=5) = \frac{2}{3} * (\frac{1}{3}) ^4 + (\frac{1}{3})^5 = \frac{3}{243}

ii)



E(X)=∑xp(x)E(X) = \sum x p(x)

1∗23+2∗29+3∗227+4∗281+5∗32431 *\frac{2}{3} + 2 * \frac{2}{9} + 3*\frac{2}{27}+4*\frac{2}{81}+ 5*\frac{3}{243}

E(X)=1.4938E(X)=1.4938


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