Probability Ques 131
Minimum number of times a fair coin must be tossed so that the probability of getting atleast one head is more than $99 \%$ is
(2019 Main 10 April II)
(a) $8$
(b) $6$
(c) $7$
(d) $5$
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Answer:
Correct Answer: 131.(c)
Solution:
Formula:
- As we know probability of getting a head on a toss of a fair coin is $P(H)=\frac{1}{2}=p$ (let)
Now, let $n$ be the minimum numbers of toss required to get at least one head, then required probability $=1-$ (probability that on all ’ $n$ ’ toss we are getting tail)
$ =1-(\frac{1}{2})^{n} \quad [\because P(\text { tail })=P(\text { Head })=\frac{1}{2}] $
According to the question,
$ 1-(\frac{1}{2})^{n} >\frac{99}{100} \Rightarrow (\frac{1}{2})^{n}<1-\frac{99}{100} $
$\Rightarrow \quad (\frac{1}{2})^{n} <\frac{1}{100} \Rightarrow 2^{n}>100 $
$\Rightarrow \quad n =7 \quad \text { [for minimum] }$
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