MCQ
Two different coins are tossed simultaneously. The probability of getting at least one head is:
- A$\frac{1}{4}$
- B$\frac{1}{8}$
- ✓$\frac{3}{4}$
- D$\frac{7}{8}$
✓
Answer
Correct option: C.
$\frac{3}{4}$
When two different coins are tossed simultaneously, then total possibilities = 4,
i.e. (H, H), (H, T), (T, H), (T, T)
Number of favourable outcomes for at least one head = 3, i.e. (H, T), (T, H), (T, H).
$\therefore$ Probability of getting at least one nead $=\frac{3}{4}$
i.e. (H, H), (H, T), (T, H), (T, T)
Number of favourable outcomes for at least one head = 3, i.e. (H, T), (T, H), (T, H).
$\therefore$ Probability of getting at least one nead $=\frac{3}{4}$
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Ram and Shyam said the following :
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Shyam : $8 b^2=9 a c$
Which of them is/are correct?