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The probability that a person has 0, 1, 2, 3, 4, or 5 or more siblings is 0.15, 0.49, 0.27, 0.06, 0.02, 0.01, respectively.

a) Three friends who are not siblings are gathered. What is the probability that they combined have three siblings?

b) Emma and Jacob are not siblings, but combined they have a total of 3 siblings. What is the probability that Emma has no siblings?

So for a) I have calculated that there are 7 possible ways this can occur: 1 situation where all three friends have 1 sibling, 3 cases where two friends have zero siblings and the other one has 3 siblings, and 3 cases where one person has 1 sibling, one person has 2 siblings and the last person has 0 siblings. So calculating this I obtain:

Using the notation that $P(X = N)$ means the probability that one person has $N$ siblings.

$$P(X = 1)^3 + 3*P(X = 0)^2 * P(X = 3) + 3*P(X = 2)*P(X = 1) * P(X = 0) \approx 0.181$$ Is this correct?

For part b) I used the notation $P(E = N)$ to mean the probability that Emma has N siblings, and similarly $P(J = N)$ to mean the probability that Jacob has $N$ siblings.

I interpreted the problem as asking us to calculate $$P(E = 0 | E + J = 3) = \frac{P (E = 0 \cap E + J = 3)}{P(E + J = 3)} = \frac{P(E+J = 3| E = 0) * P(E = 0)}{P(E+J = 3)} = \frac{P(J = 3)*P(E = 0)}{2*P(E = 0)*P(J = 3) + 2 * P(E = 1)*P(J = 2)} \approx 0.4416$$ where I have used Baye's rule to transform the expression.

Are my calculations correct?

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2 Answers 2

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So for a) I have calculated that there are 7 possible ways this can occur

I calculated 10 possible ways thus I did not check your calculations

  • $0-0-3$

  • $0-1-2$

  • $0-2-1$

  • $0-3-0$

  • $1-0-2$

  • $1-1-1$

  • $1-2-0$

  • $2-0-1$

  • $2-1-0$

  • $3-0-0$

Summing the single probabilities I get $0.240769\approx 24\%$

For (b) I get

$$\frac{0.15\times0.06}{0.15\times0.06+0.49\times0.27+0.27\times0.49+0.06\times0.15}=0.031847\approx 3\%$$

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In first part of the question, when you are counting cases for $2, 1 $ and $0$ siblings, there are $3! = 6$ ways, instead of $3$. They are $\{0, 1, 2\}, \{0, 2, 1\}, \{1, 0, 2\}, \{1, 2, 0\}, \{2, 0, 1\}, \{2, 1, 0\}$. So the probability should be

$P(X = 1)^3 + 3 \ P(X = 0)^2 \ P(X = 3) + 6 \ P(X = 2) \ P(X = 1) \ P(X = 0) \approx 0.2408$

For the second one, you have written the formula correctly but the answer seems wrong.

It should be $ = \frac{0.06 \times 0.15}{2 \times 0.06 \times 0.15 + 2 \times 0.49 \times 0.27} \approx 0.0318$

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