A card is drawn at random from a pack of $52$. According to the value of this card ($A =1, J=Q=K=10$) as many further cards are drawn. What is the probability that the ace of hearts is among those drawn(including the first card)? [Answer = $5/34$]
I tried splitting it up into 2 parts:
Part 1 for the ace: It can either be the ace of hearts at first which is $(1/4)(51/52)$ or ace of hearts second ${^4}C_3\cdot 1/51$. But I can't seem to get the answer please help.
Best Answer
Splitting the problem is a good idea.
1) A heart is drawn first. There a two cases
a) Ace of hearts: $\frac1{52} \quad (\color{blue}{I})$
We are done.
b) An ace of not hearts and then the ace of hearts: $\frac3{52}\cdot \frac1{51} \quad (\color{blue}{II})$
2)
We can draw the number $2$ ($k=2$) and then draw two cards where one of them is the ace of hearts. The probability is
$$\frac4{52}\cdot \frac{\binom{1}{1}\cdot \binom{50}{1}}{\binom{51}{2}}$$
We can sum up the probabilities from $k=2$ to $k=9$
$$\frac4{52}\cdot \left( \sum_{k=2}^9\frac{\binom{1}{1}\cdot \binom{50}{k-1}}{\binom{51}{k}}\right) \quad (\color{blue}{III})$$
3) What is left where we have to draw 10 after the first draw. This is the case when we have drawn a $10$, Jack, Queen or King.
We have $k=10$ and regard the $4$ cases by multiplying it by $4$
$$4\cdot \frac4{52}\cdot \frac{\binom{1}{1}\cdot \binom{50}{9}}{\binom{51}{10}} \quad (\color{blue}{IV})$$
What is left is to sum up the intermediate results: $ (\color{blue}{I})+ (\color{blue}{II})+(\color{blue}{III})+(\color{blue}{IV})$
Especially for the third term a calculator is needed. Here is what I got as the final result. It is the proposed solution.