Question 854713: 7 slips of papers with names of the seven days of the week are placed in a hat. Two slips of
paper are drawn out without replacement; these will be the days of classes.
What is the probability that no classes is scheduled at the weekend?
Found 2 solutions by ewatrrr, Theo:
Answer by ewatrrr(24785)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! 7 slips of paper, each slip references one day of the week.
the weekend therefore is contained in two of these slips.
the probability that you will get 0 classes on the weekend is 5/7 * 4/6 = 20/42 = 10/21.
this is because 5 out of the 7 days are not on the weekend.
the number of ways you can get 2 slips of paper out of 7 is equal to 7C2 which is equal to (7*6) / 2 which is equal to 21.
the number of ways you can get 2 out of 5 days that are not weekends is 5C2 which is equal to 5*4 / 2 which is equal to 20 / 2 which is equal to 10.
the probability is therefore equal to 5C2 / 7C2 which is equal to 10/21 that you will not get a weekend day.
this is small enough that i can show you all the different arrangement.
let the letters a and b represent the weekend days.
let the number 1,2,3,4,5 represent the weekday days.
the pot contains a,b,1,2,3,4,5
you draw 2 slips of paper at a time.
the possible combinations are:
ab *****
a1 *****
a2 *****
a3 *****
a4 *****
a5 *****
b1 *****
b2 *****
b3 *****
b4 *****
b5 *****
12
13
14
15
23
24
25
34
35
45
out of the 21 possible pairs, 11 of them contain a weekend day.
this means there are 10 pairs that do not contain a weekend day.
the probability that you will not have a class on the weekend is 10/21.
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