Question

Ram is known to hit a target into 2 out of 3 whereas sham is know is known to hit the target in 5 out of 11 shots what is the probability that the target is hit ?? ​

Answer 1

SOLUTION

GIVEN

Ram is known to hit a target into 2 out of 3 whereas sham is known to hit the target in 5 out of 11 shots

TO DETERMINE

The probability that the target is hit

EVALUATION

Let A be the event that the target is hitted by Ram and B be the event that the the target is hitted by Sham

Then it is given that Ram is known to hit a target into 2 out of 3

$\therefore \: \: \displaystyle \sf{P(A) = \frac{2}{3} }$

Again it is also stated that Sham is known to hit the target in 5 out of 11 shots

$\therefore \: \: \displaystyle \sf{P(B) = \frac{5}{11} }$

Now the required probability that the target is hit

$\displaystyle \sf{P(A \cup \: B)}$

$= \displaystyle \sf{P(A) +P(B) - P(A \cap \: B)}$

$\displaystyle \sf{ = P(A) +P(B) - P(A).P(B)} \: \: ( \because \: A \: and \: B \: are \: independent \: )$

$\displaystyle \sf{ = \frac{2}{3} + \frac{5}{11} - \bigg( \frac{2}{3} \times \frac{5}{11} \bigg)}$

$\displaystyle \sf{ = \frac{2}{3} + \frac{5}{11} - \frac{10}{33} }$

$\displaystyle \sf{ = \frac{22 + 15 - 10}{33} }$

$\displaystyle \sf{ = \frac{27 }{33} }$

$\displaystyle \sf{ = \frac{9 }{11} }$

FINAL ANSWER

$\displaystyle \sf{The \: probability \: that \: the \: target \: is \: hit \: = \frac{9 }{11} }$

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