You have a bag with 3 red, 3 white, and 2 blue balls in it. You draw three balls without replacement. What is the probability that the 3rd ball chosen is blue?
So you want the sequence
XXB , that is, notblue-notblue-blue
prob = (6/8)(5/7)(2/6) = 5/28
no, the first or 2nd ball may also be blue. In fact, they may both be blue, giving 0 probability of 3rd blue in that case.
ok, I misread the question.
All Cases , where X is not blue, 2 blue, 6 non-blue
BBB --- not possible, only 2 blue
BBX
BXB ---> (2/8)(6/7)(1/6) = 1/28
XBB ---> (6/8)(2/7)(1/6) = 1/28
BXX
XBX
XXB ---> (6/8)(5/7)(2/6) = 5/28
XXX
your prob = 7/28
To find the probability that the 3rd ball chosen is blue, we need to consider the total number of possible outcomes and the number of favorable outcomes.
Step 1: Find the total number of possible outcomes.
When drawing three balls without replacement, the first ball has 8 options, the second ball has 7 options (since one ball has already been drawn), and the third ball has 6 options (since two balls have already been drawn).
Therefore, the total number of possible outcomes is given by:
Total number of outcomes = 8 * 7 * 6 = 336.
Step 2: Find the number of favorable outcomes.
We want to get a blue ball on the third draw. There are 2 blue balls remaining in the bag when we draw the third ball.
Since we already selected 2 balls and there are 3 red and 3 white balls remaining, we can calculate the number of ways to choose the first two balls (which will be non-blue) using the combination formula:
Number of ways to choose 2 non-blue balls = C(6, 2) = 15.
So, the number of favorable outcomes, in this case, is 2.
Step 3: Calculate the probability.
Using the formula for probability: probability = number of favorable outcomes / total number of possible outcomes, we can substitute in the values we found in steps 1 and 2:
Probability = 2 / 336 = 1 / 168.
Therefore, the probability that the 3rd ball chosen is blue is 1/168.