IPMAT Indore 2023 · Quantitative Ability · Question 37

Step 1 — use the complement. “At least one head” is the opposite of “no heads at all”, and the all-tails outcome has a clean probability:

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

Step 2 — impose the condition.

1-\frac{1}{{2}^{n}}>0.8 \Rightarrow \frac{1}{{2}^{n}}<0.2 \Rightarrow{2}^{n}>5

Step 3 — take the smallest such n.

{2}^{2}=4 \left( \text{not}>5 \right),{2}^{3}=8>5 ✓

n=3

Step 4 — verify. With 3 tosses, P=1-1/8=0.875>0.8 ✔ With 2 tosses, P=1-1/4=0.75, which does not exceed 0.8 ✔

Answer: Option 1 — 3

Shortcut. “At least one” always means “one minus none”. That converts the question to {2}^{n}>5, and since {2}^{3}=8 is the first power of 2 above 5, the answer is 3.

Do not be talked out of a small answer. Three of the four options are 5, 6 and 7, which makes 3 look suspiciously easy — but 0.875>0.8 settles it. Sanity-check by computing the probability for your chosen n and for n-1; it takes five seconds and removes all doubt.

The same one-line method covers the whole family: for P\left( \text{at least one}\right)>p with a fair coin, solve {2}^{n}>\frac{1}{1-p}.