JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Probability XII

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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

Answer: 8

Solution

Let the coin be tossed

n

times. The probability of getting at least two heads is:

P (at least 2 heads) = 1 - P (0 heads) - P (1 head)

= 1 - (\frac{1}{2})^{n} - n (\frac{1}{2})^{n} = 1 - \frac{n + 1}{2 ^{n}}

We need:

1 - \frac{n + 1}{2 ^{n}} \geq 0.96

\frac{n + 1}{2 ^{n}} \leq 0.04 = \frac{1}{25}

\frac{2 ^{n}}{n + 1} \geq 25

Testing values:

$n = 7$ : $\frac{128}{8} = 16 < 25$
$n = 8$ : $\frac{256}{9} \approx 28.4 > 25$ ✓

The minimum value is

n = 8

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Integer answer

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