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The soccer team\'s shirts have arrived in a big box, and people just start grabb

ID: 3201204 • Letter: T

Question

The soccer team's shirts have arrived in a big box, and people just start grabbing them, looking for the right size. The box contains 6 medium, 9 large, and 6 extra large shirts. You want a medium for you and one for your sister. Find the probability for each event described.
a) The first two you grab are the wrong sizes.
b) The first medium shirt you find is the third you one you check.
c) The first four shirts you pick are all extra-large.
d) At least one of the four shirts you check is a medium.

Explanation / Answer

(a)

Given:

Medium = 6

Large = 9

Extra Large = 6

Total = 21

(a) First 2 are wrong. i.e., both of them are not medium.

This is not binomial distribution, because the item is not replaced after each trial.

First trial:

Number of favourable events = 15 (15 are not medium)

Total number of events = 22

Probability of first trial not medium = 15/22.

Second trial:

Number of favourable events = 14 ( 1 non medium is taken in trial 1)

Total number of events = 21 (1 shirt is taken in trial 1)

Probability of second trial not medium = 14/21

Probability of first trial not medium AND Probability of second trial not medium = 15/22 X 14/21 = 0.4545

(b) First not medium.

P(not M) = 15/22

Second not medium = 14/21

Third trial - medium

Number of favourable events = 6 (No medium is taken in trial 1 & 2. So, Medium remains same 6)

Total number of events = 20 (2 shirts are taken in trial 1 & 2)

P(M) = 6/20

Probability of first trial not medium AND second trial not medium AND third trial medium = 15/22 X 14/21 X 6/20 = 0.1364

(c) First trial:

P(extra large) = 6/22

Second trial:

P(extra large) = 5/21

Third trial:

P(extra large) = 4/20

Fourth trial:

P(extra large) = 3/19

P(4 medium) = 6/22 X 5/21 X 4/20 X 3/19 = 0.0010

(d) P(atleast 1 of 4 is a M) = 1 - P(all the 4 not M)

Trial 1:

P(not M) = 15/22

Trial 2 :

P(not M)= 14/21

Trial 3:

Trial 4:

P(not M) = 12/19

P(all 4 not M)=15/22 X 14/21 X 13/20 X 12/19 = 0.1866.

P(atleast 1 of 4 is M) = 1 - 0.1866 = 0.8134