Question

The Jones family was one of the first to come to the U.S. They
had 5 children. Assuming that the probability of a child being a
girl is .5, find the probability that the Jones family had:

at least 2 girls?

at most 4 girls?

Answer #1

a)

Here, n = 5, p = 0.5, (1 - p) = 0.5 and x = 2

As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)

We need to calculate P(X >= 2).

P(X >= 2) = (5C2 * 0.5^2 * 0.5^3) + (5C3 * 0.5^3 * 0.5^2) + (5C4
* 0.5^4 * 0.5^1) + (5C5 * 0.5^5 * 0.5^0)

P(X >= 2) = 0.3125 + 0.3125 + 0.1563 + 0.0313

P(X >= 2) = 0.8126

b)

Here, n = 5, p = 0.5, (1 - p) = 0.5 and x = 4

As per binomial distribution formula P(X = x) = nCx * p^x * (1 -
p)^(n - x)

We need to calculate P(X <= 4).

P(X <= 4) = (5C0 * 0.5^0 * 0.5^5) + (5C1 * 0.5^1 * 0.5^4) + (5C2
* 0.5^2 * 0.5^3) + (5C3 * 0.5^3 * 0.5^2) + (5C4 * 0.5^4 *
0.5^1)

P(X <= 4) = 0.0313 + 0.1563 + 0.3125 + 0.3125 + 0.1563

P(X <= 4) = 0.9689

The Jones family was one of the first to come to the U.S. They
had 5 children. Assuming that the probability of a child being a
girl is .5, find the probability that the Jones family had:
at least 2 girls?
at most 2 girls?

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had 6 children. Assuming that the probability of a child being a
girl is .5, find the probability that the Jones family had: at
least 3 girls? at most 5 girls?

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round to 4 place decimal accuracy

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