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)=?

  • $\begingroup$ Are you able to solve now? $\endgroup$ – 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.


  • $\begingroup$ 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 $\endgroup$ – TTomi Apr 29 at 13:33
  • $\begingroup$ 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. $\endgroup$ – Vizag Apr 29 at 13:59
  • $\begingroup$ Were you able to solve? $\endgroup$ – Vizag Apr 29 at 18:58
  • $\begingroup$ 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! $\endgroup$ – TTomi Apr 29 at 19:51
  • 1
    $\begingroup$ Thank you, you are great! $\endgroup$ – TTomi Apr 29 at 20:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.