1
$\begingroup$

I have this exercise which i have completed parts of like probability of red first etc but the 3 parts below I can't get for sure.

Note: Replacement doesn't matter and each ball draw is independent of another.

$\texttt{You have 5 boxes each of which contains a red ball}$, $\texttt{a blue ball, a green ball, a yellow ball and a white ball.}$ $\texttt{If you draw a ball at random from each box what is the probability that}$

(a) the last ball is red given that the first ball is red

(b) there are exactly two balls of the same colour

(c) there are no balls of the same colour

$\endgroup$
  • $\begingroup$ Is this with or without replacement? $\endgroup$ – XYZT Apr 14 '17 at 18:59
  • $\begingroup$ without replacement $\endgroup$ – Declan Lunney Apr 14 '17 at 19:03
  • $\begingroup$ Do you only draw five balls in total, one from each box? Because in that case, replacement doesn't matter and each ball draw is independent of another. $\endgroup$ – XYZT Apr 14 '17 at 19:05
  • 2
    $\begingroup$ We are glad to know you have an exercise as HW, and we hope you will work on it before passing the buck. $\endgroup$ – mlc Apr 14 '17 at 19:10
0
$\begingroup$

$(a)$ The last ball is red given that the first ball is red

If we think about this, the probability of the last ball being red is not affected by what color the first ball is. Assuming that each box contains one ball of each color, the probability of the last ball is red is $\frac{1}{5}$

$(c)$ There are no balls of the same color

This one is also fairly intuitive. Since the process of each draw is independent we can think of this probability as $\frac{5}{5}* \frac{4}{5} *\frac{3}{5} *\frac{2}{5} * \frac{1}{5}$

Let's think about each of these as each individual draw. For the 1st draw, any color you draw does not eliminate the possibility of all draws being different. Say you draw a red ball.

On the second draw, $4$ of the $5$ possibilities satisfy the condition. (As long as you do not draw red, there are not yet any balls of the same color) Assume you draw a blue ball on the second draw.

For the $3rd$ draw, there are $3$ possible colors that satisfy this condition. Hopefully, you can see where this is going.

$\endgroup$

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.