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Probability question

2015 · Shift 1 · Q32
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  5. /2015 · Shift 1 · Q32

Probability question

2015 · Shift 1 · Q32

JEE AdvancedMathematicsProbabilityNumerical+4 / −1
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 0.96,0.96,0.96, is
Numerical answer
View written solutionFree

Correct answer: 8

  1. Let the number of tosses be nnn.

  2. We want P(at least 2 heads)≥0.96.P(\text{at least 2 heads}) \ge 0.96.P(at least 2 heads)≥0.96. Using the complement, P(at least 2 heads)=1−P(0 head)−P(1 head).P(\text{at least 2 heads})=1-P(0\text{ head})-P(1\text{ head}).P(at least 2 heads)=1−P(0 head)−P(1 head).

  3. For a fair coin, P(0 head)=(12)n,P(0\text{ head})=\left(\frac12\right)^n,P(0 head)=(21​)n, and P(1 head)=(n1)(12)n=n(12)n.P(1\text{ head})=\binom{n}{1}\left(\frac12\right)^n=n\left(\frac12\right)^n.P(1 head)=(1n​)(21​)n=n(21​)n.

So, P(at least 2 heads)=1−1+n2n.P(\text{at least 2 heads})=1-\frac{1+n}{2^n}.P(at least 2 heads)=1−2n1+n​.

  1. Now impose the condition: 1−n+12n≥0.96.1-\frac{n+1}{2^n} \ge 0.96.1−2nn+1​≥0.96. Thus, n+12n≤0.04=125.\frac{n+1}{2^n} \le 0.04=\frac{1}{25}.2nn+1​≤0.04=251​. So we need 25(n+1)≤2n.25(n+1) \le 2^n.25(n+1)≤2n.

  2. Check small integer values of nnn:

  • For n=7n=7n=7: P(at least 2 heads)=1−8128=1−0.0625=0.9375<0.96.P(\text{at least 2 heads})=1-\frac{8}{128}=1-0.0625=0.9375<0.96.P(at least 2 heads)=1−1288​=1−0.0625=0.9375<0.96. So n=7n=7n=7 does not work.

  • For n=8n=8n=8: P(at least 2 heads)=1−9256=247256≈0.96484>0.96.P(\text{at least 2 heads})=1-\frac{9}{256}=\frac{247}{256}\approx 0.96484>0.96.P(at least 2 heads)=1−2569​=256247​≈0.96484>0.96. So n=8n=8n=8 works.

Therefore, the minimum number of tosses required is 8.\boxed{8}.8​.

  1. Comparison with stored answer: The stored correct answer is 888, which matches the derived answer.
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