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Probability

Question
CBSEENMA12033915
Wired Faculty App

The probability of shooter hitting a target is 3 over 4. How many minmum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?

Solution

Let the shooter fire n times. 
        Now,    straight p space equals space 1 fourth comma space space straight q space equals space 1 space minus space straight p space space equals space 1 minus 1 fourth space equals space 3 over 4
                straight P left parenthesis straight X space equals space straight x right parenthesis space equals space straight C presuperscript straight n subscript straight x space straight q to the power of straight n minus straight x end exponent space straight p to the power of straight x space equals space straight C presuperscript straight n subscript straight x space open parentheses 1 fourth close parentheses to the power of straight n minus straight x end exponent space open parentheses 3 over 4 close parentheses to the power of straight x space equals space straight C presuperscript straight n subscript straight x space 3 to the power of straight x over 4 to the power of straight n.
Now, P(hitting the target at least once) > 0.99                     (given)
therefore space space space space space space straight P left parenthesis straight x space greater or equal than space 1 right parenthesis space greater or equal than space 0.99 rightwards double arrow space space space space space 1 space minus space straight P left parenthesis straight x space equals space 0 right parenthesis space greater than space 0.99 therefore space space space space space 1 space minus space straight C presuperscript straight n subscript 0 space 1 over 4 to the power of straight n greater than space 0.99 therefore space space space space space space space space space space space space space space space space space 1 over 4 to the power of straight n space less than space 0.001 space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space space space straight C presuperscript straight n subscript 0 space equals space 1 right square bracket
or         4 to the power of straight n space greater than space fraction numerator 1 over denominator 0.01 end fraction space equals space 100                                  ...(1)
The minimum value of n to satisfy the inequality (1) is 4. 
∴   the shooter must fire 4 times.

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