0
$\begingroup$

We have $2$ black and $3$ white balls in a bag, and a dice. We role a dice, then we (randomly) pick a ball from a bag $n$-times (we note a colour of a ball, then we return it to bag), where $n$ - number of dots on a dice that showed up after a throw (assume that dice is a standard one). Calculate the probability, that the $n$ is equal $5$, if we pulled out a black ball three times. Calculate the probability, that we pulled out the white ball $4$ times.

We need to use Bayes' theorem here, but I have no idea how.

Edit. My attempt:

$A_5$ - we got $n=5$ dots from throwing a dice

$B_3$ - we got $3$ black balls from a bag

$\mathbb{P}\left(A_5 | B_3\right)=\frac{\mathbb{P}\left(A_5\right)\cdot\mathbb{P}\left(B_3|A_5\right)}{\mathbb{P}\left(B_3\right)}$

And now I have to find, let's say $\mathbb{P}\left(B_3\right)$

$\mathbb{P}\left(B_3\right)=??$

  • 0
    It's slightly hard to know how to answer this without knowing what you *do* understand about the problem. – 2017-01-30
  • 0
    @SlugPue I've edited my post – 2017-01-30

1 Answers 1

1

You are off to a good start. To find $\mathbb P (B_3)$ you typically use the law of total probability, like this:

$$ \mathbb P(B_3) = \sum^{6}_{i = 1} \mathbb P(B_3 \mid A_i)\mathbb P(A_i) $$

You probably see that for example $\mathbb P(B_3 \mid A_1) = 0$ since you cannot draw $3$ balls when $n=1$. You have to, however, calculate $\mathbb P(B_3 \mid A_i)$ for $i \geq 3$.

As to $\mathbb P(B_3 \mid A_5)$, this is the equivalent of 3 successes in 5 trials, which has a binomial distribution.