Question

A box ha 3 netball and 6 volleyball card. What is the probability of selecting ....

A. A netball card, keeping it out, and then selecting another netball card?

B. A netball card, keeping it out, and then selecting another volleyball card?

Answer 1

A. $P={\frac 3 {3+6}}*{\frac {3-1} {3+6-1}}={\frac 1 3}*{\frac 1 4}={\frac 1 {12}}$

B. $P={\frac 3 {3+6}}*{\frac 6 {3+6-1}}={\frac 1 3}*{\frac 3 4}={\frac 1 4}$