Multiplying by 4

‘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)

Hello, 12345678!

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.

Click to expand...

**

You found the probability of getting odd-odd-odd-even in that order.

But we can get 3 odds in four ways: .\(\displaystyle \begin{Bmatrix}odd-odd-odd-even \\ odd-odd-even-odd \\ odd-even-odd-odd \\ even-odd-odd-odd \end{Bmatrix}\)

12345678 said:

‘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’

Click to expand...

\(\displaystyle \dfrac{\binom{5}{3}(4)+\binom{5}{4}}{\binom{9}{4}}\)

See calculation here.

soroban said:

Hello, 12345678!


**

You found the probability of getting odd-odd-odd-even in that order.

But we can get 3 odds in four ways: .\(\displaystyle \begin{Bmatrix}odd-odd-odd-even \\ odd-odd-even-odd \\ odd-even-odd-odd \\ even-odd-odd-odd \end{Bmatrix}\)

Click to expand...

Ahh thanks!!