# probability(french card)

From a standard 52-card deck of French playing cards, two random cards are missing(we don't know which ones). We pick 4 cards(without replacement). 'A' is such an event, where we pick exactly 3 aces. P(A)=?

• Are you able to solve now? – Vizag Apr 28 at 19:44

We'll need to define a few events of the cards missing in order to approach this problem.

Let's define the three possible events of the cards missing first:

1. $$0$$ aces may be missing. Let's call this event $$X$$.

2. $$1$$ ace may be missing. Let's call this event $$Y$$.

3. $$2$$ aces may be missing. Let's call this event $$Z$$.

Note the importance of defining the above events in the context of our problem.

Now let's jump into the calculations:

Note the following:

$$P(A) = P(A \cap X) + P(A \cap Y) + P(A\cap Z)$$

Now use the fact that

$$P(A \cap B) = P(A|B)\times P(B)$$

Leaving out the details for you to work em out.

Cheers!

• Thank you Vizag! P(A∩Z)=0 P(A∩Y)=3C3*47C1 P(A∩X)=4C3*46C1 I am not sure whether these are correct. The result should be 0.000738, but I get a different result. What do you think? Thank you – TTomi Apr 29 at 13:33
• You forgot two things in the above comment. 1.) You forgot to multiply by the $P(Y)$, $P(X)$ in the last two expressions respectively. Read the answer again and note that $P(A\cap B) = P(A|B)\times P(B)$. So for example $P(A \cap Y) = P(A|Y)*P(Y)$ where $P(Y) = \frac{\binom{4}{1}\times \binom {48}{1}}{\binom{52}{2}}$. And 2.) I think you forgot to add the denominator in the comment above. Make these corrections and repost a comment and I'll check whether they are right. – Vizag Apr 29 at 13:59
• Were you able to solve? – Vizag Apr 29 at 18:58
• Thank you, you are totally right, I forgot the denominators in the rush. My new solution is: 𝑃(𝐴∩𝑌)=𝑃(𝐴|𝑌)∗𝑃(𝑌) = ((3C3*47C1)/(50C2)) * (4C1*48C1)/(52C2) 𝑃(𝐴∩X)=𝑃(𝐴|X)∗𝑃(X) = ((4C3*46C1)/(50C2)) * (4C0*48C2)/(52C2) 𝑃(𝐴∩Z)=0 Thank you! – TTomi Apr 29 at 19:51
• Thank you, you are great! – TTomi Apr 29 at 20:10