0
$\begingroup$

A bag contains $10$ blue marbles, $20$ green marbles and $30$ red marbles. A marble is drawn from the bag, its color recorded and it is put back in the bag. This process is repeated $3$ times. The probability that no two of the marbles drawn have the same color is

  1. $1/36$
  2. $1/6$
  3. $1/4$
  4. $1/3$

Somewhere it explained as :

No two marbles have the same color means, the final outcome of the three draws must be a permutation of

Blue, Green, Red

There are $3! = 6$ such permutations possible.

Now, probability of getting a Blue first, Green second and Red third $= 10/60 * 20/60 *30/60$

Required probability$= 6 * 10/60 * 20/60 * 30/60 = 1/6$


But I'm not getting above explanation.

Can you explain little bit please?

$\endgroup$
5
  • 2
    $\begingroup$ It is not easy to "explain an explanation". This especially if it is a good explanation. Could you describe your problem with it? $\endgroup$
    – drhab
    Commented Nov 21, 2015 at 9:12
  • $\begingroup$ @drhab, You can add your solution please, if you know that. $\endgroup$ Commented Nov 21, 2015 at 9:13
  • 2
    $\begingroup$ Probability on $BGR$ (in that order) is $\frac16\frac26\frac36$. Do you agree with that? Likewise you can find the probabilities for other orders like $GBR$. They give the same probability and there are $3!=6$ of them. These events are disjoint and their union is the event: "no two of the marbles have the same color". So a multiplication with $6$ gives the final answer. $\endgroup$
    – drhab
    Commented Nov 21, 2015 at 9:17
  • $\begingroup$ Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. $\endgroup$
    – Did
    Commented Nov 21, 2015 at 9:21
  • $\begingroup$ yes, it's possible of any permutation of $3$. Thanks, I got it. $\endgroup$ Commented Nov 21, 2015 at 9:22

1 Answer 1

2
$\begingroup$

If the marbles are drawn in another order, we have a different event. This is the reason why we have to multiply the probability with $3!=6$.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .