AP EAMCET · Maths · Straight Lines
6 coins are tossed 320 times. The probability of getting 5 heads 2 times is
- A \(30^2 \times \frac{e^{-30}}{2}\)
- B \(30 \times e^{-30}\)
- C \(30^2 \times e^{-30}\)
- D \(30 \times e^{-10}\)
Answer & Solution
Correct Answer
(A) \(30^2 \times \frac{e^{-30}}{2}\)
Step-by-step Solution
Detailed explanation
\( p(\text{5 heads}) = C(6, 5) (1/2)^5 (1/2)^1 = 6 \times (1/2)^6 = 6/64 = 3/32 \) \( \lambda = \text{number of trials} \times p = 320 \times (3/32) = 30 \) \( P(X=2) = \frac{e^{-\lambda} \lambda^2}{2!} = \frac{e^{-30} 30^2}{2} = 30^2 \times \frac{e^{-30}}{2} \)
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