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If I launch 4 dice and at least one even number appears, what is the probability that the sum of the results be even?

A: "The sum of the results is even"
B: "At least one even number appears"

A∩B = A

f(x) = (4Cx)*[(1/2)^x]*[(1/2)^(n-x)]

P(A) = f(2)+f(4) = [(4C2)*[(1/2)^4]]+[(1/2)^4]
P(B) = 1-f(0) = 1-[(1/2)^4]

P(A/B) = P(A)/P(B) = (7/16)/(15/16) = 7/15

The solution is correct?

  • 1
    When you say "pair" do you mean "two the same" or "a multiple of $2$"? – 2017-01-21
  • 0
    I mean a multiple of 2. – 2017-01-21

2 Answers 2

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Your answer is correct, but your work has some errors, as listed below (which work out so as to cancel each other).

  • $A \cap B$ is not equal to $A$, since $B$ is not a subset of $A$.
  • $P(A)$ is equal to $f(0) + f(2) + f(4)$, not $f(2) + f(4)$.
  • The formula for $P(A\mid B)$ should be $\displaystyle{\frac{P(A \cap B)}{P(B)}}$, not $\displaystyle{\frac{P(A)}{P(B)}}$.

Here is a corrected version, using your approach ...

$P(A \cap B) = f(2) + f(4) = 7/16$.

$P(B) = 1 - f(0) = 15/16$.

$P(A\mid B) = \displaystyle{\frac{P(A \cap B)}{P(B)}} = \frac{(7/16)}{(15/16)} = 7/15$.

1

You are correct, though I would do the calculation slightly differently: each die can be odd or even with probability $\frac12$, so there are $16$ equally probable results, $8$ of which have even sums.

We should exclude the case where all four dice are odd: odd + odd + odd + odd = even, so the conditional probability is $\dfrac{8-1}{16-1}=\dfrac{7}{15}$

  • 0
    How do you know that half the total number of cases adds an even number, without seeing the set A? If I launch the die a odd number of times, would it change? If the die had an odd number of faces, would it change? Thank you. – 2017-01-21
  • 0
    @CarlosFrostte: the last die has a probability $\frac12$ of being odd or even, and this mean the total has a probability $\frac12$ of being odd or even. Throwing an odd number of dice would not change this, though it would change other parts of the calculation since the final result would become greater than $\frac12$. Having dice with an odd number of faces would change the calculations substantially, since you would no longer have half the faces being even – 2017-01-21