# Finding the probability that a team roster contains Ted and Vanessa

A school Science Fair team will consist of 2 male students, two female students and one alternate. The alternate can be of either sex. There are $4$ males (including Ted) and $5$ females (including Vanessa) trying out for the team. A team roster is a list of five students chosen without mention of which one of the five will be the alternate. If the five students are chosen randomly from among the nine trying out, find the probability that the team roster contains Ted and Vanessa.

I tried solving it by finding the situations where we pick 2 males out of 3 and 1 female out of 4 and 1 male out of 3 and 2 females out of 4 to find the possible outcomes and divided them with the total number of outcomes.

So i got $(3*4+3*6)/300$ which seems to be wrong as the answer is 3/10

• What have you attempted? Where are you stuck? – N. F. Taussig Oct 11 '17 at 2:16
• Please edit your question to show how you got your answer. That will make it easier to see where you made your error. – N. F. Taussig Oct 11 '17 at 2:18
• i edited the question to show what i got – Dhruv Raghunath Oct 11 '17 at 2:23
• Your numerator is correct. You made your error in the denominator. – N. F. Taussig Oct 11 '17 at 2:27

A team roster consists of three males and two females or two males and three females. Since there are four male and five female students available, the number of possible team rosters is $$\binom{4}{3}\binom{5}{2} + \binom{4}{2}\binom{5}{3}$$ As you observed, if both Ted and Vanessa are selected, a team with three males and two females can be obtained by selecting two of the other three males and one of the other four females, and a team with two males and three females can be obtained by selecting one of the other three males and two of the other four females. Hence, the number of favorable cases is $$\binom{3}{2}\binom{4}{1} + \binom{3}{1}\binom{4}{2}$$ Therefore, the desired probability is $$\frac{\dbinom{3}{2}\dbinom{4}{1} + \dbinom{3}{1}\dbinom{4}{2}}{\dbinom{4}{3}\dbinom{5}{2} + \dbinom{4}{2}\dbinom{5}{3}}$$