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Probability

Question
CBSEENMA12033921

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Solution
Let n be the number of times a fair coin is tossed.
         Now            straight p space space equals space 1 half comma space space straight q space equals space 1 minus straight p space equals space 1 minus 1 half space equals space 1 half
              P(at least one head) > 90%                         (given)
rightwards double arrow space space space 1 minus space straight P left parenthesis 0 right parenthesis thin space greater than space 90 over 100 space space space space space space rightwards double arrow space space space space space 1 space minus space straight C presuperscript straight n subscript 0 space straight q to the power of straight n space equals space 9 over 10
rightwards double arrow space space 1 minus space space 1 space cross times space open parentheses 1 half close parentheses to the power of straight n space greater than space 9 over 10 space space space space space space space rightwards double arrow space space space space 1 minus space 9 over 10 greater than 1 over 2 to the power of straight n
rightwards double arrow space space space space space space space space 1 over 10 greater than 1 over 2 to the power of straight n space space space space space space space space space space space space space rightwards double arrow space space space space 2 to the power of straight n space greater than space 10
    
∴ least value of n is 4
∴  minimum number of tosses is 4.

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