There are 5 wrong questions in a question paper, consisting of 10 questions. If 6 questions are selected from this question paper, then the probability that there will be exactly 3 wrong questions among them is
To find the probability, we need to determine the total number of ways to choose 6 questions from the question paper and the number of ways to choose exactly 3 wrong questions among them.
1. Total number of ways to choose 6 questions from a set of 10 questions can be calculated using the combination formula: C(n, r) = n! / (r!(n-r)!), where n is the total number of questions and r is the number of questions to be chosen.
So, the total number of ways to choose 6 questions from 10 is C(10, 6) = 10! / (6!(10-6)!) = 210.
2. Now, let's determine the number of ways to choose exactly 3 wrong questions among the 6 chosen questions.
Since there are 5 wrong questions in the question paper, we need to choose 3 of those wrong questions and 3 from the remaining 5 correct questions.
The number of ways to choose 3 wrong questions from 5 is C(5, 3) = 5! / (3!(5-3)!) = 10.
Similarly, the number of ways to choose 3 correct questions from 5 is C(5, 3) = 5! / (3!(5-3)!) = 10.
3. To determine the probability, we divide the number of favorable outcomes (number of ways to choose 3 wrong questions and 3 correct questions) by the total number of possible outcomes (number of ways to choose 6 questions from 10).
So, the probability = (Number of ways to choose 3 wrong questions) * (Number of ways to choose 3 correct questions) / (Number of ways to choose 6 questions from 10)
= 10 * 10 / 210
= 100 / 210
= 5 / 21.
Therefore, the probability that there will be exactly 3 wrong questions among the 6 chosen questions is 5/21.
prob of wrong = 5/10 = 1/2
Prob(exactly 3 wrong when selecting 6)
= (6,3)(1/2)^3 (1/2)^3
= 20(1/64) = 5/16