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

[zorro]

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 ?)}

[Martingale]

...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

[zorro]

12 is not a probability

Could u please help me solve this problem by pointing out the mistake

[stapel_eliz]

Assuming the decimal points stand for multiplication, could you please show the steps in your arithmetic with the fractions?

[zorro]

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

[Martingale]

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.