Question

Box i contains 4 red and 8 blue marbles while box ii contains 5 red and 3 blue marbles. an unfair coin is tossed –whose probability of turning up heads is 40%. if the coin comes up heads box i is chosen and a random marble is chosen, otherwise if it is tails the marble is chosen from box ii.(a)find the probability a red marble is chosen.(b)if a red marble is chosen, what is the probability it came from box i? (c)if a blue marble is chosen, what is the probability it came from box i?

Answer 1

Denote by $B_1,B_2$ the event that a marble is drawn from box 1 or 2, respectively, and by $R$ the event that a red marble is drawn. If a blue marble is drawn, we'll denote that by the complement of $R$, or $R^C$.

a. If a red marble is drawn, either it's drawn from box 1 or box 2. The events $B_1$ and $B_2$ are mutually exclusive, so we have

$\mathbb P(R)=\mathbb P((R\cap B_1)\cup(R\cap B_2))=\mathbb P(R\cap B_1)+\mathbb P(R\cap B_2)$

Now invoke the definition of conditional probability:

$\mathbb P(R)=\mathbb P(B_1)\mathbb P(R\mid B_1)+\mathbb P(B_2)\mathbb P(R\mid B_2)$

We're explicitly given the probability of drawing from either box. We also know the probabilities of drawing a red marble from box 1 or 2 are, respectively,

$\mathbb P(R\mid B_1)=\dfrac{\dbinom41\dbinom80}{\dbinom{12}1}=\dfrac4{12}=\dfrac13$

and

$\mathbb P(R\mid B_2)=\dfrac{\dbinom51\dbinom30}{\dbinom81}=\dfrac58$

So

$\mathbb P(R)=\dfrac25\times\dfrac13+\dfrac35\times\dfrac58=\dfrac{61}{120}\approx0.508$

b. Given that we draw a red marble, we're looking for the probability that it was drawn from box 1, i.e. $\mathbb P(B_1\mid R)$. We use Bayes' theorem:

$\mathbb P(B_1\mid R)=\dfrac{\mathbb P(B_1\cap R)}{\mathbb P(R)}=\dfrac{\mathbb P(B_1)\mathbb P(R\mid B_1)}{\mathbb P(R)}$

which is really just a matter of using the definition of conditional probability to rewrite the numerator on the the right side, as

$\mathbb P(R\cap B_1)=\mathbb P(R)\mathbb P(B_1\mid R)=\mathbb P(B_1)\mathbb P(R\mid B_1)$

We know $\mathbb P(R)$ from part (a), so we compute

$\mathbb P(B_1\mid R)=\dfrac{\dfrac25\times\dfrac13}{\dfrac{61}{120}}=\dfrac{16}{61}\approx0.262$

c. Now we look for $\mathbb P(B_1\mid R^C)$, which can be computed similarly.

$\mathbb P(B_1\mid R^C)=\dfrac{\mathbb P(B_1\cap R^C)}{\mathbb P(R^C)}=\dfrac{\mathbb P(B_1)\mathbb P(R^C\mid B_1)}{1-\mathbb P(R)}$
$\mathbb P(B_1\mid R^C)=\dfrac{\dfrac25\times\dfrac23}{1-\dfrac{61}{120}}=\dfrac{32}{59}\approx0.542$