Probability 5 Question 3
3. 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: 3. (c)
Solution:
- 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,
$$ \begin{array}{rlrl} & 1-\frac{1}{2}^{n} & >\frac{99}{100} \Rightarrow \frac{1}{2}^{n}<1-\frac{99}{100} \\ \Rightarrow \quad \frac{1}{2}^{n} & <\frac{1}{100} \Rightarrow 2^{n}>100 \\ \Rightarrow \quad n & =7 \quad \text { [for minimum] } \end{array} $$
