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Number 9 a, b, and c is what I need help with please

Number 9 a, b, and c is what I need help with please-example-1
User Exussum
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a) The probability that none of the 5 women will have twins is approximately 0.5905.

b) The probability that exactly 1 of the 5 women will have twins is approximately 0.0027.

c) The probability that at least 3 of the 5 women will have twins is approximately 0.0005.

a) Probability of None Having Twins

To calculate the probability that none of the 5 women will have twins, we use the probability of not having twins for one woman (1 − 1/10) and raise it to the power of 5 (since the events are independent).

P(none having twins) = (1 − 1/10)⁵

P(none having twins) ≈ 0.3487

b) Probability of Exactly 1 Having Twins

The probability of exactly 1 woman having twins is found by combining the probability of one woman having twins (1/10) with the probability of the remaining women not having twins.

P(exactly 1 having twins) = (⁵P₁)(1/10)¹(9/10)⁴

P(exactly 1 having twins) = (0.0417)(0.1)¹(0.9)⁴

P(exactly 1 having twins) = (0.0417)(0.1)(0.6561)

P(exactly 1 having twins) ≈ 0. 0027

c) Probability of at Least 3 Having Twins

To find the probability that at least 3 women will have twins, we sum the probabilities of exactly 3, 4, and 5 women having twins.

P(at least 3 having twins) = ∑⁵ₙ₌₃[(⁵Pₙ)(1/10)ⁿ(9/10)⁽⁵⁻ⁿ⁾]

P(at least 3 having twins) = (0.5)(0.001)(0.81) + (1)(0.0001)(0.9) + (1)(0.00001)(1)

P(at least 3 having twins) ≈ 0.000505

Therefore, probability that none of the 5 women will have twins is approximately 0.5905, probability that exactly 1 of the 5 women will have twins is approximately 0. 0027 and probability that at least 3 of the 5 women will have twins is approximately 0.0005.

User Michaelrbock
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