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The outcome of a random variable X is

$X = \{0, 1, 2, 3, 4\}$

with associated probabilities

$P\left(X\right)=\{0.6, 0.2, 0.1, 0.05, 0.05\}$

I would like to compute

$P\left(x=0|x \leq3 \right)$

I recall that the general identity for a conditional probability involving two events is

$P\left ( x = 0 | x \leq 3 \right ) = \frac{P\left ( x = 0 \,\cap \, x \leq 3 \right )}{P\left(x \leq 3 \right )}$

and

$P\left(x \leq 3\right) = 0.95$

Any hints to further my understanding is appreciated.

Thanks in advance

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  • $\begingroup$ There is an error in your Bayes formula. The denominator should correspond to the condition, i.e. P(x<= 3) instead of P(x=0). Now if you figure out the numerator, then you are done. Hint: What is the intersection of A and B if A is a subset of B? $\endgroup$
    – LiKao
    Commented Aug 24, 2018 at 11:54
  • $\begingroup$ @LiKao It is just A itself. Would you be willing to post your hint as an answer? I will accept your answer and provide an upvote $\endgroup$
    – Physkid
    Commented Aug 24, 2018 at 12:03

1 Answer 1

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There is an error in your Bayes formula. The denominator should correspond to the condition, i.e. P(x<= 3) instead of P(x=0). Thus your formula should read (before edit):

$P(x=0|x\leq3)=\frac{P(x=0\cap x\leq3)}{P(x\leq 3)}$

Now if you figure out the numerator, then you are done.

Hint: What is the intersection of A and B if A is a subset of B?

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