Pramod goes to a shop to buy some clothes. Shopkeeper shows him 7 shirts, 5 pants and 10 t-shirts. If he selects three clothes at random, then find the probability that, at least one of the clothes is shirt. (Enter the answer correct to two decimal places)
Question
Pramod goes to a shop to buy some clothes. Shopkeeper shows him 7 shirts, 5 pants and 10 t-shirts. If he selects three clothes at random, then find the probability that, at least one of the clothes is shirt. (Enter the answer correct to two decimal places)
Solution
To solve this problem, we first need to calculate the total number of ways Pramod can select three clothes from the given options. This is a combination problem, so we use the combination formula: C(n, r) = n! / [r!(n-r)!], where n is the total number of items, r is the number of items to choose, and "!" denotes factorial.
Step 1: Calculate total number of ways to select 3 clothes The total number of clothes is 7 shirts + 5 pants + 10 t-shirts = 22 clothes. So, the total number of ways to select 3 clothes out of 22 is C(22, 3) = 22! / [3!(22-3)!] = 1540 ways.
Step 2: Calculate the number of ways to select 3 clothes with no shirts The total number of clothes excluding shirts is 5 pants + 10 t-shirts = 15 clothes. So, the number of ways to select 3 clothes out of 15 (with no shirts) is C(15, 3) = 15! / [3!(15-3)!] = 455 ways.
Step 3: Calculate the probability of selecting at least one shirt The probability of an event is the number of favorable outcomes divided by the total number of outcomes. Here, the favorable outcome is selecting at least one shirt, which is the total ways of selecting 3 clothes minus the ways of selecting 3 clothes with no shirts. So, the probability = (Total ways - Ways with no shirts) / Total ways = (1540 - 455) / 1540 = 1.085 / 1.540 = 0.7045.
So, the probability that at least one of the clothes is a shirt is approximately 0.70 or 70% when rounded to two decimal places.
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