Knowing that in the first 3 draws without replacement from a deck of 52 french cards (26 red|26 black) at least one red card has been drawn, what is the probability that the 2nd draw has been a red card?
How would you solve this problem?
What’s the solution to this conditional probability problem
conditional probabilityprobability
Best Answer
At the first 3 cards at least one red card has been drawn.
1) 2 black & 1 red cards
2) 1 black & 2 red cards
3) 3 red cards
We don't know it is case 1, 2, or 3.
We have to consider all the possibilities.
Case 1) First, what is the probability that we are in case 1? It is important, because case 1 happens much more frequently than the other cases, we have to give a higher weight. And then calculate the probability the next one is red under the condition of case 1.
Pr(case1) = $ \frac{{26 \choose 2}{26 \choose 1}}{52 \choose 3}$
This is not the probability I want to use. What I want to use is the probability of case 1 under the condition of at least 1 red card drawn.
Therefore Pr(case1|at least 1 red) = Pr(case 1| Case1 or Case 2 or Case3)
$=\frac{\text{Pr(case1)}}{\text{Pr(case 1)+Pr(case 2)+Pr(case 3)}}$
Pr(red|case1) = $\frac{25}{49} $
where, 25 is the number of red cards remain under the case 1. And 49 is the number of total cards.
Pr(red & case 1|at least 1 red) = Pr(red|case1)xPr(case1|At least 1 red)
And repeat the calculation for cases 2 and 3.
Pr(red|Case1 or case2 or case3) = Pr(red & case 1|at least 1 red) + Pr(red & case 2|at least 1 red) + Pr(red & case 3| at least 1 red)