Please help me to answer this conditional probability question

Question

Question:

So, for part (a), my answer is:

$$ {10 \choose 8} (0.6)^8 (0.4)^2 + {10 \choose 9}(0.6)^9 (0.4)^1 + {10 \choose 10 } (0.6)^10 (0.4)^0 = 0.1673$$

I am not sure how to answer part (b).

What I thought of:

Let $X$ = number of correct answers $Y$ = number of answers known

Therefore:

$P(X\ge 9|Y\ge 8) = \frac{P\left(X\ge9, Y=8 \right) + P\left(X\ge9, Y=9 \right) + P\left( X\ge 9 , Y=10 \right)}{P\left( Y\ge 8 \right)}$

In this case, the denominator is the answer to part (a).

I am not sure if I got the numerator right. I think I enumerated all the possible cases but now when I want to compute the first term in the numerator, I need to (I think) compute:

Probability know any 8 correct answers and guess any 1 answer correctly and any 1 answer wrong + Probability know any 8 correct answer and guess 2 correctly.

I am not sure how to compute the above and I am not even sure if this is the way to proceed

Answer 1

Hint:

$$P(X≥9, Y≥8) = P(X=9, Y=8) + P(X=10, Y=8) + P(X=9, Y=9) + P(X=10, Y=9) + P(X=10, Y=10)$$

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How to calculate $P(X=9, Y=8)$?

• I knew 8 answers: the probability for that is $\left(\frac{3}{5}\right)^8$
• I guessed one answer correct: the probability for that is $\frac{2}{5}\frac{1}{3}=\frac{2}{15}$
• I guessed one answer incorrect and didn't knew it: the probability
  for that is $1-\frac{3}{5}-\frac{2}{5}\frac{1}{3}=\frac{4}{15}$.

There are $\frac{10!}{8!\cdot 1!\cdot 1!}$ ways to order these answers, so:

$$P(X=9, Y=8) = \frac{10!}{8!\cdot 1!\cdot 1!}\cdot \left(\frac{3}{5}\right)^8 \cdot \frac{2}{15} \cdot \frac{4}{15}$$