SMC2UCLA2_ said:
Ok. here it is.
The probability of a boy is 1/2 which is also the probability of a girl.
There are several ways to get 3 boys and 1 Girl and in this question the order doesnt matter so we are looking at the following combinations: GBBB, BGBB, BBGB, and BBBG (B= prob of boy and G = prob of girl).
You can find the probability of each of these seperately and each will = 1/16. (1/2 x 1/2 x 1/2 x 1/2)
Your final answer will be P(GBBB) + P(BGBB) + P (BBGB) + P(BBBG) = 1/16 + 1/16 + 1/16 + 1/16 = 4/16 = 1/4.
OR... you can use Binomial Expansion to determine the probability of a particular combination rather than going through all possibilities.
(a + b)1 = a + b
(a + b)2 = a2 + 2ab + b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6
Notice the coefficients for each term come from pascals triangle. In this example we would be looking at (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.
lets let a = prob of girl and b = prob of a boy and a = b = 1/2.
According to the binomial expansion, the probability of 3 boys and 1 girl will be 4a3b where 3 is an exponent.
So we get.... 4(1/2)^3 (1/2) = 4/16 = 1/4.