0
$\begingroup$

Two boxes, the first contains 3 red & 2 white balls. The second box contains 4 red & 7 white balls.

A fair die is tossed. If the outcome is even, a ball is drawn from the first box, while a ball is drawn from the second box if the outcome is odd.

Find the probability that a randomly chosen ball will be white and it comes from the first box.

Attempt at solution:

P(drawing from first box)= 3/6= 1/2, P(drawing a white ball)=9/16, P(White ball ∩ first box)= P(White ball | first box)xP(First box) =(2/5)(1/2)=1/5,

My friend says that the required probability is not P( White ball ∩ first box) but instead it's P(White ball | first box), which is it?

$\endgroup$
  • $\begingroup$ It is just a matter of reading the question correctly ("...white and it comes from the first box), and you did that. Your answer is correct. Your friend makes it: "...white if it comes from the first box". $\endgroup$ – drhab Oct 17 '15 at 18:43
  • $\begingroup$ Your interpretation is correct. $\endgroup$ – true blue anil Oct 17 '15 at 19:16
1
$\begingroup$

For questions about conditional probability, usually the word given will be in the phrase somewhere.

  • Find the probability that a randomly chosen ball will be white and it comes from the first box.
  • Find the probability that a randomly chosen ball will come from the first box and be white.

both represent $P(\text{White}\cap \text{First box})$


  • Find the probability that a randomly chosen white ball comes from the first box.
  • Find the probability that a randomly chosen ball comes from the first box given that it is white.

represents $P(\text{First box}~|~\text{White})$


  • Find the probability that a randomly chosen ball from the first box is white.
  • Find the probability that a randomly chosen ball is white given that it comes from the first box.

represents $P(\text{White}~|~\text{First box})$


I agree with your interpretation and that the probability being searched for is indeed $P(\text{White}\cap \text{First box}) = \frac{1}{5}$

$\endgroup$
1
$\begingroup$

The probability that the ball gets chosen from the first box is $\frac{1}{2}$, and the proability that a ball chosen from the first box is white is $\frac{2}{5}$.

The probability that a ball gets chosen from the first box and that it is white is $\frac{1}{2} \times \frac{2}{5} = \frac{1}{5}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.