Probability 5 Question 18
18. The minimum number of times a fair coin needs to be tossed, so that the probability of getting atleast two heads is atleast 0.96 , is
(2015 Adv.)
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Answer:
Correct Answer: 18. (8)
Solution:
- Using Binomial distribution,
$$ \begin{aligned} P(X \geq 2) & =1-P(X=0)-P(X=1) \\ & =1-\frac{1}{2}{ }^{n}-{ }^{n} C _1 \cdot \frac{1}{2} \cdot \frac{1}{2}{ }^{n-1} \\ & =1-\frac{1}{2^{n}}-{ }^{n} C _1 \cdot \frac{1}{2^{n}}=1-\frac{1+n}{2^{n}} \end{aligned} $$
Given, $P(X \geq 2) \geq 0.96$
$$ \begin{aligned} & \therefore \quad 1-\frac{(n+1)}{2^{n}} \geq \frac{24}{25} \\ & \Rightarrow \quad \frac{n+1}{2^{n}} \leq \frac{1}{25} \\ & \therefore \quad n=8 \end{aligned} $$
