In a drawer Richard has 5 pairs of gloves. Each pair has a different color. On Day 1 Richard selects two individual gloves at random from the 10 gloves in the drawer. On Day 2 Richard selects 2 of the remaining 8 gloves at random and on Day 3 two of the remaining 6 gloves at random. The probability that Day 3 is the first day Richard selects matching gloves is m/n, where m and n are relatively prime positive integers. What ism+n?

To find the probability that Day 3 is the first day Richard selects matching gloves, we need to calculate the probability of not selecting matching gloves on Day 1 and Day 2, and then calculate the probability of selecting matching gloves on Day 3.

On Day 1, Richard selects 2 individual gloves at random from the 10 gloves in the drawer. The probability of not selecting matching gloves on Day 1 can be calculated as:

P(not matching on Day 1) = (number of ways to select 2 non-matching gloves) / (total number of ways to select 2 gloves)

There are 10 gloves to choose from, so there are 10C2 = 45 total ways to select 2 gloves. Since we want non-matching gloves, we can choose any one of the 5 pairs of gloves (10 gloves) and then any one of the remaining 8 individual gloves. Therefore, there are 5*8 = 40 ways to select 2 non-matching gloves.

P(not matching on Day 1) = 40 / 45 = 8/9

On Day 2, Richard selects 2 of the remaining 8 gloves at random. The probability of not selecting matching gloves on Day 2 can be calculated similarly:

P(not matching on Day 2) = (number of ways to select 2 non-matching gloves) / (total number of ways to select 2 gloves)

Now there are only 8 gloves to choose from, so there are 8C2 = 28 total ways to select 2 gloves. Since we want non-matching gloves, we can choose any one of the remaining 4 pairs of gloves (8 gloves) and then any one of the remaining 6 individual gloves. Therefore, there are 4*6 = 24 ways to select 2 non-matching gloves.

P(not matching on Day 2) = 24 / 28 = 6/7

On Day 3, Richard selects 2 of the remaining 6 gloves at random. The probability of selecting matching gloves on Day 3 (given that he did not select matching gloves on Day 1 and Day 2) can be calculated as:

P(matching on Day 3) = (number of ways to select 2 matching gloves) / (total number of ways to select 2 gloves)

Now there are only 6 gloves to choose from, so there are 6C2 = 15 total ways to select 2 gloves. Since we want matching gloves, we can choose any one of the remaining 3 pairs of gloves (6 gloves). Therefore, there are 3 ways to select 2 matching gloves.

P(matching on Day 3) = 3 / 15 = 1/5

To find the overall probability that Day 3 is the first day Richard selects matching gloves, we multiply the probabilities from each day:

P(Day 3 is the first day) = P(not matching on Day 1) * P(not matching on Day 2) * P(matching on Day 3) = (8/9) * (6/7) * (1/5) = 48/315

Therefore, m = 48 and n = 315.

m + n = 48 + 315 = 363

The value of m+n is 363.

To find the probability that Day 3 is the first day Richard selects matching gloves, we can consider the possible outcomes on each day.

On Day 1, there are 10 gloves in the drawer, and Richard randomly selects 2 gloves. The probability of selecting any particular pair of matching gloves is 1/10 (since there is only one pair among the 10 gloves). Therefore, the probability of not selecting a matching pair on Day 1 is 1 - 1/10 = 9/10.

On Day 2, there are 8 gloves remaining in the drawer, and Richard randomly selects 2 gloves. The probability of selecting any particular pair of matching gloves is now 1/8. However, we also need to consider the probability of not selecting a matching pair on Day 1. Since this event and the events of Day 2 are independent, we can multiply the probabilities. Therefore, the probability of not selecting a matching pair on Day 1 and 2 is (9/10) * (1 - 1/8) = (9/10) * (7/8) = 63/80.

On Day 3, there are 6 gloves remaining in the drawer, and Richard again randomly selects 2 gloves. The probability of selecting any particular pair of matching gloves is now 1/6. To find the probability that Day 3 is the first day Richard selects matching gloves, we need to consider the probabilities of not selecting a matching pair on Day 1 and 2. Again, since these events are independent, we can multiply the probabilities. Therefore, the probability of not selecting a matching pair on Day 1, 2, and 3 is (9/10) * (7/8) * (1 - 1/6) = (9/10) * (7/8) * (5/6) = 315/480 = 21/32.

Finally, the probability that Day 3 is the first day Richard selects matching gloves is 1 - 21/32 = 11/32.

Therefore, m = 11 and n = 32, and m+n = 11 + 32 = 43.