Probability Ques 132
The minimum number of times one has to toss a fair coin so that the probability of observing atleast one head is atleast $90 \%$ is
(2019 Main, 8 April II)
(a) $2$
(b) $3$
(c) $5$
(d) $4$
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Answer:
Correct Answer: 132.(d)
Solution:
- The required probability of observing atleast one head
$=1-P(\text { no head }) $
$=1-\frac{1}{2^{n}} \quad[\text { let number of toss are } n] $
$[\because P(\text { Head })=P(\text { Tail })=\frac{1}{2}]$
According to the question, $1-\frac{1}{2^{n}} \geq \frac{90}{100}$
$\Rightarrow \frac{1}{2^{n}} \leq \frac{1}{10} \Rightarrow 2^{n} \geq 10 \Rightarrow n \geq 4$
So, minimum number of times one has to toss a fair coin so that the probability of observing atleast one head is atleast $90 \%$ is $4$ .
BETA
