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Two coins are tossed simultaneously. Find the probability of getting at least one head. |
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Answer» Given that two coins are tossed simultaneously. Hence the possible outcomes are {(H, H), (H, T), (T, H), (T, T)} which are equally likely. Hence, total possible outcomes = n(S) = 4. Let the event E be getting at least one head. Therefore E = {(H, H), (H, T), (T, H)}, Hence the number of outcomes favourable to E is n(E) = 3. \(\therefore\) P(E) = \(\frac{n(E)}{n(S)} = \frac{3}{4}\). Hence, the probability of getting at least one head is \(\frac{3}{4}\). |
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