SATHEE: Probability 5 Question 3

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

(d) 5

Show Answer

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)

$= 1 - {\frac{1}{2}}^{n} ∵ P (tail) = P (Head) = \frac{1}{2}$

According to the question,

$\begin{aligned} 1 - {\frac{1}{2}}^{n} & > \frac{99}{100} \Rightarrow {\frac{1}{2}}^{n} < 1 - \frac{99}{100} \\ \Rightarrow {\frac{1}{2}}^{n} & < \frac{1}{100} \Rightarrow 2^{n} > 100 \\ \Rightarrow n & = 7 [for minimum] \end{aligned}$

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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

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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