In a bag of 12 marbles are ‘n’ red marbles and the remaining are blue marbles. If Josh takes 2 marbles from the bag, what is the probability that he takes 1 red marble and 1 blue marble?
Question
In a bag of 12 marbles are ‘n’ red marbles and the remaining are blue marbles. If Josh takes 2 marbles from the bag, what is the probability that he takes 1 red marble and 1 blue marble?
Solution
To solve this problem, we need to understand that the probability of an event is the number of ways that event can occur divided by the total number of outcomes.
Step 1: Identify the total number of outcomes. Josh is taking 2 marbles from a bag of 12. The total number of ways to do this is given by 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. So, the total number of outcomes is C(12, 2) = 12! / [2!(12-2)!] = 66.
Step 2: Identify the number of ways the event of interest can occur. The event of interest is that Josh takes 1 red marble and 1 blue marble. The number of ways this can occur is the product of the number of ways to choose 1 red marble and the number of ways to choose 1 blue marble. This is given by C(n, 1) * C(12-n, 1) = n * (12-n).
Step 3: Calculate the probability. The probability is the number of ways the event of interest can occur divided by the total number of outcomes. So, the probability is [n * (12-n)] / 66.
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