Two integers are selected at random from integers 1 to 11. If the sum ...
Out of integers from 1 to 11, there are 5 even and 6 odd integers.
Let A – both the numbers chosen are odd
B –sum of the numbers chosen is even.
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Two integers are selected at random from integers 1 to 11. If the sum ...
Understanding the Problem:
When two integers are selected randomly from the set {1, 2, 3, ..., 11}, we need to find the probability that both numbers are odd given that their sum is even.
Identifying the Possible Scenarios:
- There are a total of 11 integers in the set {1, 2, 3, ..., 11}.
- Out of these, there are 6 odd numbers: 1, 3, 5, 7, 9, 11.
- There are 5 even numbers: 2, 4, 6, 8, 10.
Finding the Probability:
- When two numbers are selected randomly, there are a total of 11C2 = 55 ways to choose them.
- For the sum to be even, both numbers must be either odd or even.
- The number of ways to choose 2 odd numbers = 6C2 = 15 ways.
- The number of ways to choose 2 even numbers = 5C2 = 10 ways.
- Therefore, the total number of favorable outcomes = 15 ways (both numbers are odd).
Calculating the Probability:
- Probability = (Number of favorable outcomes) / (Total number of outcomes)
- Probability = 15 / 55 = 3 / 11.
Therefore, the probability that both numbers are odd given that their sum is even is 3/11. Hence, option 'C' (3/5) is the correct answer.
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