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.