A box contains four red, three blue, and six green lenses. Two lenses are randomly selected from the box. Find the probability of getting two red lenses when two are selected if the first selection is NOT replaced before the second selection is made.
Pr=4/(13)*3/12=1/13
To find the probability of getting two red lenses without replacement, we need to consider the number of ways we can select two red lenses out of the total possible combinations of selecting any two lenses from the box.
Total number of lenses in the box = 4 (red) + 3 (blue) + 6 (green) = 13 lenses
Let's calculate the probability step-by-step:
Step 1: Calculate the number of ways to choose two red lenses out of four.
We can use the binomial coefficient formula to calculate this: C(n, r) = n! / (r! * (n-r)!)
C(4, 2) = 4! / (2! * (4-2)!)
= 4! / (2! * 2!)
= (4 * 3 * 2 * 1) / (2 * 1 * 2 * 1)
= 6
There are 6 ways to choose two red lenses out of the 4 available.
Step 2: Calculate the number of possible combinations to choose any two lenses from the box.
We can use the binomial coefficient formula to calculate this as well.
C(13, 2) = 13! / (2! * (13-2)!)
= 13! / (2! * 11!)
= (13 * 12 * 11!)/ (2 * 1 * 11!)
= (13 * 12) / (2 * 1)
= 78
There are 78 possible combinations to choose any two lenses from the box.
Step 3: Calculate the probability of getting two red lenses.
The probability is given by the number of favorable outcomes (getting two red lenses) divided by the total number of possible outcomes (choosing any two lenses).
Probability of getting two red lenses = number of ways to choose two red lenses / number of possible combinations
= 6 / 78
= 1/13
Therefore, the probability of getting two red lenses when two are selected without replacement is 1/13.
To find the probability of getting two red lenses when two are selected without replacement, we need to use the concept of conditional probability.
First, let's find the probability of selecting a red lens on the first draw. There are 4 red lenses out of a total of 13 lenses in the box, so the probability of selecting a red lens on the first draw is 4/13.
Now, after selecting a red lens on the first draw, there are 3 red lenses remaining out of a total of 12 lenses in the box. So, the probability of selecting a red lens on the second draw, given that a red lens was already selected on the first draw, is 3/12.
To find the probability of both events happening (selecting a red lens on the first draw and then a red lens on the second draw), we multiply the probabilities together:
(4/13) * (3/12) = 12/156 = 1/13.
Therefore, the probability of getting two red lenses when two are selected without replacement is 1/13.