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Three numbers are chosen at random without replacement from (1, 2, 3 ..., 10). The probability that the minimum number is 3 or the maximum number is 7 is
- a)12/37
- b)11/40
- c)13/35
- d)14/35
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Three numbers are chosen at random without replacement from (1, 2, 3 ....
Most Upvoted Answer
Three numbers are chosen at random without replacement from (1, 2, 3 ....
Given information:
Three numbers are chosen at random without replacement from (1, 2, 3 ..., 10).
To find:
The probability that the minimum number is 3 or the maximum number is 7.
Solution:
Step 1:
Total number of outcomes:
When three numbers are chosen from (1, 2, 3 ..., 10) without replacement, the total number of outcomes is given by the combination formula:
nCr = n! / (r!(n-r)!)
Here, n = 10 (total numbers to choose from)
r = 3 (number of numbers chosen)
Total number of outcomes = 10C3 = 10! / (3!(10-3)!) = 10! / (3!7!) = (10*9*8*7!)/(3*2*1*7!) = (10*9*8)/(3*2*1) = 120
Hence, there are 120 possible outcomes.
Step 2:
Favorable outcomes:
To find the favorable outcomes, we need to consider two cases separately:
Case 1: The minimum number is 3
Case 2: The maximum number is 7
Case 1: The minimum number is 3
If the minimum number is 3, then we can choose the other two numbers from the remaining 9 numbers (4, 5, 6, 7, 8, 9, 10). The number of ways to choose 2 numbers from 9 is given by the combination formula:
9C2 = 9! / (2!(9-2)!) = 9! / (2!7!) = (9*8)/(2*1) = 36
Case 2: The maximum number is 7
If the maximum number is 7, then we can choose the other two numbers from the remaining 9 numbers (1, 2, 3, 4, 5, 6). The number of ways to choose 2 numbers from 6 is given by the combination formula:
6C2 = 6! / (2!(6-2)!) = 6! / (2!4!) = (6*5)/(2*1) = 15
Total favorable outcomes:
Total favorable outcomes = favorable outcomes in Case 1 + favorable outcomes in Case 2 = 36 + 15 = 51
Step 3:
Probability:
Probability = (Total favorable outcomes) / (Total number of outcomes) = 51 / 120 = 17 / 40
Therefore, the probability that the minimum number is 3 or the maximum number is 7 is 17/40, which is equivalent to option B.
Three numbers are chosen at random without replacement from (1, 2, 3 ..., 10).
To find:
The probability that the minimum number is 3 or the maximum number is 7.
Solution:
Step 1:
Total number of outcomes:
When three numbers are chosen from (1, 2, 3 ..., 10) without replacement, the total number of outcomes is given by the combination formula:
nCr = n! / (r!(n-r)!)
Here, n = 10 (total numbers to choose from)
r = 3 (number of numbers chosen)
Total number of outcomes = 10C3 = 10! / (3!(10-3)!) = 10! / (3!7!) = (10*9*8*7!)/(3*2*1*7!) = (10*9*8)/(3*2*1) = 120
Hence, there are 120 possible outcomes.
Step 2:
Favorable outcomes:
To find the favorable outcomes, we need to consider two cases separately:
Case 1: The minimum number is 3
Case 2: The maximum number is 7
Case 1: The minimum number is 3
If the minimum number is 3, then we can choose the other two numbers from the remaining 9 numbers (4, 5, 6, 7, 8, 9, 10). The number of ways to choose 2 numbers from 9 is given by the combination formula:
9C2 = 9! / (2!(9-2)!) = 9! / (2!7!) = (9*8)/(2*1) = 36
Case 2: The maximum number is 7
If the maximum number is 7, then we can choose the other two numbers from the remaining 9 numbers (1, 2, 3, 4, 5, 6). The number of ways to choose 2 numbers from 6 is given by the combination formula:
6C2 = 6! / (2!(6-2)!) = 6! / (2!4!) = (6*5)/(2*1) = 15
Total favorable outcomes:
Total favorable outcomes = favorable outcomes in Case 1 + favorable outcomes in Case 2 = 36 + 15 = 51
Step 3:
Probability:
Probability = (Total favorable outcomes) / (Total number of outcomes) = 51 / 120 = 17 / 40
Therefore, the probability that the minimum number is 3 or the maximum number is 7 is 17/40, which is equivalent to option B.
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Community Answer
Three numbers are chosen at random without replacement from (1, 2, 3 ....
Sample space =10C3 =120,
no. of ways to select 3 as minimum=7C2=21
no. of ways to select 7 as maximum=6C2=15
But 4,5,6 are being repeated in 2 cases, so favourable ways are 7C2+6C2-3C2=33
req probability=33/120=11/40
no. of ways to select 3 as minimum=7C2=21
no. of ways to select 7 as maximum=6C2=15
But 4,5,6 are being repeated in 2 cases, so favourable ways are 7C2+6C2-3C2=33
req probability=33/120=11/40
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Three numbers are chosen at random without replacement from (1, 2, 3 ..., 10). The probability that the minimum number is 3 or the maximum number is 7 isa)12/37b)11/40c)13/35d)14/35Correct answer is option 'B'. Can you explain this answer?
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