Question

From a bowl containing five red, three white, and seven blue chips, select four at random and without replacement. Compute the conditional probability of one red, zero white, and three blue chips, given that there are at least three blue chips in this sample of four chips.

Answer 1

The probability is
`(5)/(9)`

Explanation:

Let A be the event of one red, zero white, and three blue chips,

And, B is the event of at least three blue chips,

Since, A ∩ B = A (because If A happens that it is obvious that B will happen )

Thus, the conditional probability of A if B is given,

`P((A)/(B))=(P(A\cap B))/(P(B))=(P(A))/(P(B))`

Now, red chips = 5,

White chips = 3,

Blue chips = 7,

Total chips = 5 + 3 + 7 = 15

Since, the probability of one red, zero white, and three blue chips, when four chips are chosen,

`P(A)=(^5C_1* ^3C_0* ^7C_3)/(^(15)C_4)`

`=(5* 35)/(1365)`

`=(175)/(1365)`

`=(5)/(39)`

While, the probability that of at least three blue chips,

`P(B)=(^8C_1* ^7C_3+^8C_0* ^7C_4)/(^(15)C_4)`

`=(8* 35+35)/(1365)`

`=(315)/(1365)`

`=(3)/(13)`

Hence, the required conditional probability would be,

`P((A)/(B))=(5/39)/(3/13)`

`=(65)/(117)`

`=(5)/(9)`