Question

If a sex ratio of births is 49 girls to 51 boys, the probability that there will be 8 girls amongst 10 babies born on the same day in a maternity hospital

• A. ${}^{10}{C}_8\left(0.51{)}^8\right(0.49{)}^2$
• B. ${}^{10}{C}_8\left(0.49{)}^8\right(0.51{)}^2$
• C. ${}^{10}{C}_8(0.49X0.51{)}^8$
• D. ${}^{10}{C}_8(0.49X0.51{)}^2$

Answer 1

Answer: (B)

${}^{10}{C}_8\left(0.49{)}^8\right(0.51{)}^2$

Probability of having a girl$=\frac{49}{100}=0.49$

Probability of having a boy$=\frac{51}{100}=0.51$

Selection of 8 girls out of 10 babies is ${=}^{10}{C}_8$
10 babies in which 8 are girls and 2 are boys, so probability of 8 girls amongst 10 babies born ${=}^{10}{C}_8\left(0.49{)}^8\right(0.51{)}^2$