Probability?

Question

A group of six students all have different grades in stats. The median mark in the group is 68. If 2 of the 6 students are selected at random what is the probability they both have marks greater than 68?

Im stuck here. Im going over my notes trying to figure out how to do it. Any help, answers and explanations would be appreciated.

Answer 1

This is a trick question. The median means the middle number not necessarily the average. Well, if all student had a 68, the median is 68 so no students are higher and none are lower. The odds get screwed up if 2 students had the same number, or 3, or 4, or 5, or 6. So it is a trick question.

But if it is not a trick question, than Battleship guy has the right answer.

Answer 2

[Edit: The answer is NOT 1/4, as others have said. The answer _approaches_ 1/4 as the sample gets larger. For a small group of 6, the probability is rather different. Read on.]

> The median mark in the group is 68.

That means half the students have grades over 68, and half under 68. As a mnemonic device, color the 3 high-scoring students green, and color the 3 low-scoring students red.

> If 2 of the 6 students are selected at random...

There are 15 possible pairs of students (15 = 6!/(2!4!)). So figure out, out of those 15 pairs, how many consist of two "green" students.

A little thought will show that there are only three possible "green/green" pairs; namely:

G1, G2
G1, G3
G2, G3

So, the probability of a random pair being "green/green" is 3 out of 15.

Answer 3

1 / 4 is the answer

This is because there are 6 people and the median is 68. Because the number of students is even and everyone has a different number, there is not a definite person who has 68 -- 68 is the average of the third and fourth person. This means that the fourth, fifth, and sixth people all are greater than 68. 3 / 6 = 1 / 2.

So the probability that one person is greater than 68 is 1/2. The probability that two are would be 1/2 * 1/2 or 1/4.

Answer 4

the first student pick has a 3 out of 6 chance of scoring greater than 68.

assuming the first student's score is above 68, the second student selected has a 2 out of 5 chance of having a score greater than 68.

since 3/6 = 1/2 then:
1/2 * 2/5 = 1/5.

the answer is 1 out of 5 (or 20%).

Answer 5

i think my argument ought to no longer be properly received. I say the prospect is 50% enable a ??, enable b ?? and randomly go with the values for a and b. As already stated, for a ? 0, P( a < b²) = a million, that is trivial. purely truly a lot less trivial is the concept P(a < 0 ) = a million/2 and therefore P( a < b² | a ? 0) = a million and P( a < b² ) ? a million/2 Now evaluate what occurs at the same time as a > 0 For a > 0, at the same time as it really is elementary to reveal there's a non 0 threat for a finite b, the reduce, the prospect is 0. a < b² is a similar as saying 0 < a < b², bear in options we are purely gazing a > 0. If this a finite period on an unlimited line. The threat that a is component to this period is 0. P( a < b² | a > 0) = 0 As such we've a finished threat P( a < b² ) = P( a < b² | a ? 0) * P(a ? 0) + P( a < b² | a > 0) * P(a > 0) = a million * a million/2 + 0 * a million/2 = a million/2 bear in options, it really is because of the endless contraptions. no count number what style of period you draw on paper or on a pc you'll stumble on a finite threat that seems to mind-set a million. yet it really is because of the finite random form turbines on the computer and if we had this question requested with finite values there's a a answer more suitable than 50%. i do not propose to be condescending, yet please clarify why utilising the Gaussian to approximate a uniform distribution is a good theory? are not endless numbers relaxing. Cantor at the same time as mad operating with them! :)

Answer 6

6 students, 3 lower and 3 higher than median score.

P(both higher)
= n(choose 2 from higher 3) / n(choose any 2 from all 6)
= 3C2 / 6C2
= 3/15
= 1/5
= 0.2
= 20%

Answer 7

lets see. The avarage is 68, half are above and halv below.... the probability of each being above is 0,5.... 2 events... 0,5 times 0,5 =0,25...
I hope I got it right.....I always draw the tree in my mind
Above /
Below branch to A/B or A/B