Question

A hockey all-star game has the Eastern Division all-stars play against the Western Division all-stars. On the Eastern Division team there are 8 United States-born players, 14 Canadian-born players, and 2 European-born players. On the Western Division team there are 12 United States-born players, 8 Canadian-born players, and 4 European-born players. If one player is selected at random from the Eastern Division team and one player is selected at random from the Western Division team, what is the probability that neither player will be a Canadian-born player?

Answer 1

160/576

Explanation:

Did it on college board

Answer 2

`\frac {7}{36}` .

Explanation:

Let,

A = The event that a player selected at random from the Eastern Division team is not a Canadian - born player.

B = The event that a player selected at random from the Western Division team is not a Canadian - born player.

We are to find out, P(A∩B).

According to the question,

P(A∩B)

=
`P(A) * P(B)` [Since, A, B are independent]

=
`\frac {14}{8 + 14 + 2} * \frac {8}{12 + 8 + 4}`

=
`\frac {14}{24} * \frac {8}{24}`

=
`\frac {7}{12} * \frac {1}{3}`

=
`\frac {7}{36}`