Four cards are dealt at random from a pack of 8 cards numbered 1 to 8 respectively. Given that the largest number dealt was 7, calculate the conditional probability that the smallest card dealt was 3.

Anyone any ideas how to work it out?
(S1 A-level btw)

Ta, Duane 🙂

Nope

Too difficult.

I know what the capital of Bolivia is, if that helps ?

No, but that never stopped me guessing wildly.

Highest is a 7, so the 8 is irrelevant. So, how many ways to pick 4 cards from 7 = 7!/4!3! = 35.

For 3 to be the lowest only 5 cards possible (3 – 7), so 5!/4!1! = 5 ways to select 4 cards from these. But one selection is 4,5,6,7 so 3 isn’t the lowest in that group. And 3,4,5,6 isn’t eligible either (as 7 is the highest). Hence 3 combinations of cards with 3 as the lowest and 7 the highest.

hence, my answer would be 3/35.

Let the flaming begin…