Question

A box contains 2 black, 4 white and 3 red balls. One ball is drawn at random from the box, another ball is drawn at random and kept beside the first. This process is repeated till all the balls are drawn from the box. Find the probability that the balls drawn are in the sequence of 2 black, 4 white and 3 red.

Answer 1

The probability of getting 2 black balls in the first 2 draws is
\[\frac{2}{9}\times\frac{1}{8}\]
The probability of getting 4 white balls in the next 4 draws is
\[\frac{1}{7C4}=\frac{1}{\frac{7!}{4!3!}}=\frac{1}{35}\]
The probability of getting 3 red balls in the next 3 draws is 1.
Therefore the required probability is given by
\[\frac{2}{9\times8\times35}\]

Answer 2

is it correct ?

Answer 3

Your fraction is not the correct result based on my posted solution.

Answer 4

ok

Answer 5

The number of sequences is just the number of permutations of the balls, i.e. \((2+4+3)!=9!\). Now, the key is that since the order for each color ball doesn't matter, there are e.g. \(k!\) ways to pick \(k\) balls of a color in sequence so \(2!4!3!\) ways to pick the above sequence of colored balls. hence the probability of picking such a sequence is \(\dfrac{2!4!3!}{9!}=\dfrac1{1260}\approx0.08\%\)

Answer 6

this is the same result @kropot72 found but I think it is more illuminating to have multiple approaches that are equivalent :-) nice job btw @kropot72