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A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occurred in a certain town, which had 14,391 children.

a) 0.000011
b) 0.000022
c) 0.000033
d) 0.000044

User Den Pat
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1 Answer

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Final answer:

To find the probability that more than four children suffer from this rare form of malignant tumor, we need to calculate the probability of four or fewer cases and subtract it from 1.

Step-by-step explanation:

To find the probability that more than four children suffer from this rare form of malignant tumor, we need to calculate the probability of four or fewer cases and subtract it from 1.

The probability of four or fewer cases can be found using binomial probability.
Binomial probability formula: P(X ≤ k) = C(n, k) * p^k * (1 - p)^(n - k), where C(n, k) is the combination function and p is the probability of success of a single trial. In this case, p = 0.000011 and n = 14391.

Using the formula, we calculate the probability of four or fewer cases: P(X ≤ 4) = C(14391, 0) * 0.000011^0 * (1 - 0.000011)^(14391 - 0) + C(14391, 1) * 0.000011^1 * (1 - 0.000011)^(14391 - 1) + C(14391, 2) * 0.000011^2 * (1 - 0.000011)^(14391 - 2) + C(14391, 3) * 0.000011^3 * (1 - 0.000011)^(14391 - 3) + C(14391, 4) * 0.000011^4 * (1 - 0.000011)^(14391 - 4)

Subtracting this probability from 1 will give us the probability of more than four cases.

User UncleAdam
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