Probability
Balls drawn without replacement — finding N
MJAT_TS7_P2
Grade 12
Question:
A bag has $N$ balls: 3 white, 6 green, and the rest blue. Three drawn without replacement. $P(W_1\cap G_2\cap B_3)=\dfrac{52}{N}$ and $P(B_3|W_1\cap G_2)=\dfrac{9}{2}$. Then $N$ equals:
Step-by-Step Solution
Key Concept: $P(W_1\cap G_2\cap B_3)=\frac{3}{N}\cdot\frac{6}{N-1}\cdot\frac{N-9}{N-2}$. $P(B_3|W_1\cap G_2)=\frac{N-9}{N-2}=\frac{9}{2}$... hmm that gives $N-9=(9/2)(N-2)\Rightarrow 2N-18=9N-18\Rightarrow-7N=0\Rightarrow N=0$. Doesn't work. Try $P(B_3|W_1\cap G_2)=\frac{N-9}{N-2}=\frac{2}{9}$: $9(N-9)=2(N-2)\Rightarrow 9N-81=2N-4\Rightarrow 7N=77\Rightarrow N=11$.
$N=\mathbf{11}$.
Correct Answer: 11