Algebra

Probability

Medium

applied

How many minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

✅ Correct Option: 2

The probability of getting at least one head can be calculated using the complement:

P(at least one head) = 1 - P(all tails)

For a fair coin, P(tail) = $\frac{1}{2}$

If the coin is tossed $n$ times:

P(all tails) = $\left(\frac{1}{2}\right)^n$

P(at least one head) = $1 - \left(\frac{1}{2}\right)^n$

The probability must be more than 90%:

$1 - \left(\frac{1}{2}\right)^n > 0.90$

$\left(\frac{1}{2}\right)^n < 0.10$

For $n = 3$ :

$\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125$

P(at least one head) = $1 - 0.125 = 0.875 = 87.5\%$

This is less than 90%.

For $n = 4$ :

$\left(\frac{1}{2}\right)^4 = \frac{1}{16} = 0.0625$

P(at least one head) = $1 - 0.0625 = 0.9375 = 93.75\%$

This is more than 90%.

Therefore, the minimum number of tosses required is 4.

2025: 15 May Shift 1

Match List-I with List-II

if $P(A) = \frac{3}{7}$ , $P(B) = \frac{4}{7}$ and $P(A \cup B) = \frac{5}{7}$

List-I	List-II
(A) $P(A \cap B)$	(I) $\dfrac{2}{3}$
(B) $P(A \mid B)$	(II) $\dfrac{5}{7}$
(C) $P(B \mid A)$	(III) $\dfrac{1}{2}$
(D) $P(A' \cup B')$	(IV) $\dfrac{2}{7}$

Choose the correct answer from the options given below:

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How many minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90%?

Solution

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