‘A bag contains 9 discs numbered 1, 2, 3, 4, 5, 6, 7, 8, 9’.
‘Martin chooses 4 discs at random, without replacement. Find the probability that the 4 digits include at least 3 odd digits’
So I calculated getting 3 odds;
5/9 * 4/8 * 3/7 * 4/6 = 5/63
I then did 4 odds;
5/9 * 4/8 * 3/7 * 2/6 = 5/126
I then added these to get 5/42.
However, the answer is 5/14.
The mark scheme multiplies the P of getting 3 odds by 4, but not the second- I was wondering why?
(So they did (5/63 * 4) + 5/126 =5/14)
‘Martin chooses 4 discs at random, without replacement. Find the probability that the 4 digits include at least 3 odd digits’
So I calculated getting 3 odds;
5/9 * 4/8 * 3/7 * 4/6 = 5/63
I then did 4 odds;
5/9 * 4/8 * 3/7 * 2/6 = 5/126
I then added these to get 5/42.
However, the answer is 5/14.
The mark scheme multiplies the P of getting 3 odds by 4, but not the second- I was wondering why?
(So they did (5/63 * 4) + 5/126 =5/14)