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Q. 10 How many times must a man toss a fair coin so that the probability of having at least one head is more than $90^{o}/_{o}$ ?

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Let the man toss coin n times.

Probability of getting head in first toss = P

$P=\frac{1}{2}$

$q=\frac{1}{2}$

$\therefore \, \, \, \, P(X=x)=^nC_x.q^{n-x}.p^x$

$=^nC_x.\left ( \frac{1}{2} \right )^{n-x}.\left ( \frac{1}{2} \right )^x$

$P(getting\, atleast\, one\, head)> \frac{90}{100}$

$P(X\geq 1)> 0.9$

$1-P(X=0)> 0.9$

$1-^nC_0 \frac{1}{2^n}> 0.9$

$^nC_0 \frac{1}{2^n}< 0.1$

$\frac{1}{2^n}< 0.1$

$\frac{1}{0.1}< 2^n$

$10< 2^n$

The minimum value to satisfy the equation is 4.

The man should toss a coin 4 or more times.

Posted by

seema garhwal

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