7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a man (so X=1 indicates that a man achieved the highest score on the exam). Find each of the following:
How do I get p(x=3) and p(x=7)?
P(x=1)= 7/10 =.70
p(x=2)= (7/10) * (3/9) = 0.233333333
p(x=3)
p(x=7)
To find the probability that X is equal to 3, we need to consider the possible placements of the 3 women and the 7 men in the rankings.
To calculate p(x=3), we can use the concept of combinations. We need to select 3 men out of the 7 available to achieve the top 3 rankings, and arrange them appropriately. Therefore, the probability can be calculated as follows:
p(x=3) = (7C3 * 3!) / 10!
Here, 7C3 represents the number of ways to select 3 men out of 7, and 3! represents the number of ways to arrange those 3 men among themselves. The denominator, 10!, represents the total number of possible rankings.
Using the combination formula: nCr = n! / (r! * (n-r)!)
p(x=3) = (7C3 * 3!) / 10!
= (7! / (3! * (7-3)!) * 3!) / 10!
= (7! / (3! * 4!) * 3!) / 10!
= (7! / (3! * 4!)) / (10! / (10-7)!)
Simplifying further, we have:
p(x=3) = (7*6*5) / (3*2*1 * 4*3*2*1) * (10*9*8*7*6*5*4*3*2*1) / (3*2*1 * (10*9*8*7*6*5))
p(x=3) = 7/120 = 0.058333333
To find p(x=7), we can use the same logic. Since X=7 indicates that a man has achieved the lowest ranking, we need to consider the placement of the 7 men below the 3 women.
p(x=7) = (7C7 * 7!) / 10!
Here, 7C7 represents the number of ways to select all 7 men, and 7! represents the number of ways to arrange those 7 men among themselves.
Simplifying further, we have:
p(x=7) = (7C7 * 7!) / 10!
= (7! / (7! * (7-7)!) * 7!) / 10!
= (7! / (7! * 0!)) / (10! / (10-7)!)
Since 0! is equal to 1, we have:
p(x=7) = (7! / 7!) / (10! / 3!)
Simplifying further, we get:
p(x=7) = 1 / (10*9*8)
p(x=7) = 1/720 = 0.0013888889
Therefore, p(x=3) is approximately 0.0583 and p(x=7) is approximately 0.0014.