The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

Question

The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

1 - P(no head) - P(one head) â¥ 0.96 1- nC0 ((1/2))n - nC1 ((1/2))n ge 0.96 (n+1/2n) le 0.04 (n+1/2n) le (1/25) Minimum value of n is 8.

Q. The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

1 - P(no head) - P(one head) $\geq 0.96$
$1 -^{n} C_{0} (\frac{1}{2})^{n} -^{n} C_{1} (\frac{1}{2})^{n} \geq 0.96$
$\frac{n + 1}{2 ^{n}} \leq 0.04$
$\frac{n + 1}{2 ^{n}} \leq \frac{1}{25}$
Minimum value of n is 8.

Q. The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is

1 - P(no head) - P(one head) ≥0.96 1−nC0​(21​)n−nC1​(21​)n≥0.96 2nn+1​≤0.04 2nn+1​≤251​ Minimum value of n is 8.

1 - P(no head) - P(one head) $\geq 0.96$
$1 -^{n} C_{0} (\frac{1}{2})^{n} -^{n} C_{1} (\frac{1}{2})^{n} \geq 0.96$
$\frac{n + 1}{2 ^{n}} \leq 0.04$
$\frac{n + 1}{2 ^{n}} \leq \frac{1}{25}$
Minimum value of n is 8.