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### prob. of 3 red, 2 white, 1 blue (is the answer correct)

Posted: Sun Jul 04, 2010 10:27 am
Question :
A box contains 5 red balls, 4 white balls and 3 blue balls . A ball is selected at random from the box . the colour of the ball is noted and then replaced back . Find the probability that out of 6 balls selected in this manner , 3 are red, 2 are white , 1 is blue.

Solution :

Let P(R) be the probability that red ball is selected i.e. P(R) = 5/12
Let P(W) be the probability that white ball is selected i.e. P(W) = 4/12
Let P(B) be the probability that blue ball is selected i.e. P(B) = 3/12

Now the probability that out of 6 balls selected 3 are red, 2 are white , 1 is blue

= P(R).P(R). P(R) + P(W).P(W) + P(B)

= 5/12 . 5/12. 5/12 + 4/12 . 4/12 + 3/12

= 144/12

= 12 (Is this correct ?)

### Re: prob. of 3 red, 2 white, 1 blue (is the answer correct)

Posted: Sun Jul 04, 2010 2:57 pm
...Now the probability that out of 6 balls selected 3 are red, 2 are white , 1 is blue

= P(R).P(R). P(R) + P(W).P(W) + P(B)

= 5/12 . 5/12. 5/12 + 4/12 . 4/12 + 3/12

= 144/12

= 12 (Is this correct ?)
12 is not a probability

### Re: prob. of 3 red, 2 white, 1 blue (is the answer correct)

Posted: Sun Jul 04, 2010 4:00 pm
12 is not a probability

Posted: Sun Jul 04, 2010 9:53 pm
Assuming the decimal points stand for multiplication, could you please show the steps in your arithmetic with the fractions?

### Re:

Posted: Mon Jul 05, 2010 3:44 am
Assuming the decimal points stand for multiplication, could you please show the steps in your arithmetic with the fractions?
Sorry silly mistakes ....
P(R).P(R).P(R) + P(W).P(W) + P(B) = 0.4289

### Re: Re:

Posted: Mon Jul 05, 2010 4:14 am
Assuming the decimal points stand for multiplication, could you please show the steps in your arithmetic with the fractions?
Sorry silly mistakes ....
P(R).P(R).P(R) + P(W).P(W) + P(B) = 0.4289

No...this problem follows a Multinomial Distribution.