A bag has 5 red and 3 blue balls. Two balls are drawn at random without replacement. What is the probability that both are red?
5/14. First red: 5/8. Second red, given the first was red: 4/7. Multiply: 5/8 x 4/7 = 20/56 = 5/14. Using combinations: C(5,2)/C(8,2) = 10/28 = 5/14.
Explain it simply
Without replacement means the first ball stays out, so the bag changes before the second draw. There is one fewer red ball and one fewer ball overall. Multiply the probability of each step, using the updated bag for the second step.
Example
Question: 5 red, 3 blue. Draw 2 without replacement. P(both red)?
Step method:
- P(first red) = 5/8.
- P(second red | first red) = 4/7.
- P(both red) = 5/8 x 4/7 = 20/56 = 5/14.
Combination method:
- Ways to pick 2 red = C(5, 2) = 10.
- Ways to pick any 2 = C(8, 2) = 28.
- P = 10/28 = 5/14.
Related: P(one red and one blue) = (5 x 3) / 28 = 15/28. P(both blue) = C(3,2)/28 = 3/28. The three add up to (10 + 15 + 3)/28 = 1.
console.log(10 + 15 + 3); // Output: 28
console.log((5 * 4) / (8 * 7) === 10 / 28); // Output: trueWhat to say in the interview
"Without replacement, the second draw has 4 reds out of 7. So 5 over 8 times 4 over 7, which is 5 over 14. The combination count, 10 out of 28, gives the same."
Common mistakes
- Using 5/8 x 5/8, which is the with-replacement answer.
- Forgetting the order factor when counting 'one red and one blue' by the step method (red then blue, or blue then red).