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Integrals

Question
CBSEENMA12035670

How many times must a fair coin be tossed so that the probability of getting at least one head is more than 80%?

Solution

Let  p denotes the probability of getting heads.
Let q denotes the probability of getting tails.
straight p space equals space 1 half
straight q space equals space 1 minus 1 half space equals 1 half
Suppose the coin is tossed n times.
Let X denote the number of times of getting heads in n trails. 
straight P left parenthesis straight X equals straight r right parenthesis space equals space straight C presuperscript straight n subscript straight r straight p to the power of straight r straight q to the power of straight n minus straight r end exponent space equals space straight C presuperscript straight n subscript straight r open parentheses 1 half close parentheses to the power of straight r open parentheses 1 half close parentheses to the power of straight n minus straight r end exponent space equals straight C presuperscript straight n subscript straight r open parentheses 1 half close parentheses to the power of straight n comma space space straight r space equals space 0 comma 1 comma space 2... comma straight n
straight P left parenthesis straight X greater or equal than 1 right parenthesis greater than 80 over 100
rightwards double arrow space space straight P left parenthesis straight X equals 1 right parenthesis plus straight P left parenthesis straight X equals 2 right parenthesis plus..... plus left parenthesis straight X equals straight n right parenthesis greater than 80 over 1000
rightwards double arrow space straight P left parenthesis straight X equals 1 right parenthesis plus straight P left parenthesis straight X equals 2 right parenthesis plus.... plus straight P left parenthesis straight X equals straight n right parenthesis plus straight P left parenthesis straight X equals 0 right parenthesis minus straight P left parenthesis straight X equals 0 right parenthesis greater than 80 over 100
rightwards double arrow 1 minus straight P left parenthesis straight X equals 0 right parenthesis greater than 80 over 100
rightwards double arrow straight P left parenthesis straight X equals 0 right parenthesis space less than 1 fifth
rightwards double arrow straight C presuperscript straight n subscript 0 open parentheses 1 half close parentheses to the power of straight n less than 1 fifth
rightwards double arrow open parentheses 1 half close parentheses to the power of straight n less than 1 fifth
rightwards double arrow straight n space equals space 3 comma 4 comma 5......
So the fair coin should be tossed for 3 or more times for getting the required probability.

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