Three fair coins are tossed simultaneously. Find the probability of getting at most two heads.

medium 2 marks Class 10 · Mathematics · Probability

Answer

\frac{7}{8}

The correct answer is $\frac{7}{8}$ — Calculate the probability: $P (E) = \frac{n ( E )}{n ( S )} = \frac{7}{8}$ .

Step-by-step solution

List all possible outcomes in the sample space $S$ : $S = {H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T}$ . The total number of outcomes is $n (S) = 8$ .
Identify the event $E$ of getting at most two heads, which means getting $0$ , $1$ , or $2$ heads (all outcomes except $H H H$ ).
List the favorable outcomes: $E = {H H T, H T H, H T T, T H H, T H T, T T H, T T T}$ , so $n (E) = 7$ .
Calculate the probability: $P (E) = \frac{n ( E )}{n ( S )} = \frac{7}{8}$ .

Common mistake: Misunderstanding 'at most two' as 'exactly two' or 'at least two'.

Concept tested: Sample space and probability of coin tosses