STK110 TUT3 Preparation sheet memo 2023
Question 1 is based on the following information:
The ratings for a group of 210 students who take both art and mathematics are summarized in
the table below.
RATING IN ART
Above average Below average Total
′
𝑨 𝑨
Above average
RATING 53 29 82
𝑴
IN
MATHEMATICS Below average
26 102 128
𝑴′
Total 79 131 210
A student is selected at random from these students.
a. What is the probability that the student will have:
1) a rating above average in art?
𝟕𝟗
𝑷(𝑨) = ≈ 𝟎. 𝟑𝟕𝟔𝟐
𝟐𝟏𝟎
2) a rating below average in mathematics?
𝟏𝟐𝟖
𝑷(𝑴′ ) = ≈ 𝟎. 𝟔𝟎𝟗𝟓
𝟐𝟏𝟎
b. What is the probability that the student will have:
1) a rating above average in mathematics given that the student has a rating above
average in art?
𝑷(𝑴∩𝑨) 𝟓𝟑⁄ 𝟓𝟑 53
𝑷(𝑴|𝑨) = = 𝟕𝟗 𝟐𝟏𝟎 = 𝟕𝟗 ≈ 𝟎. 𝟔𝟕𝟎𝟗 or from table: 𝑷(𝑴|𝑨) = 79 ≈ 𝟎. 𝟔𝟕𝟎𝟗
𝑷(𝑨) ⁄𝟐𝟏𝟎
2) a rating above average in art given that the student has a rating above average in
mathematics?
𝑷(𝑨∩𝑴) 𝟓𝟑⁄ 𝟓𝟑 𝟓𝟑
𝑷(𝑨|𝑴) = = 𝟖𝟐 𝟐𝟏𝟎 = 𝟖𝟐 ≈ 𝟎. 𝟔𝟒𝟔𝟑 or from table: 𝑷(𝑨|𝑴) = 𝟖𝟐 ≈ 𝟎. 𝟔𝟒𝟔𝟑
𝑷(𝑴) ⁄𝟐𝟏𝟎
c. What is the probability that the student will have:
1) a rating above average in both art and mathematics?
𝟓𝟑
𝑷(𝑨 ∩ 𝑴) = ≈ 𝟎. 𝟐𝟓𝟐𝟒
𝟐𝟏𝟎
2) a rating above average in art and a rating below average in mathematics?
𝟐𝟔
𝑷(𝑨 ∩ 𝑴′ ) = ≈ 𝟎. 𝟏𝟐𝟑𝟖
𝟐𝟏𝟎
1
Question 1 is based on the following information:
The ratings for a group of 210 students who take both art and mathematics are summarized in
the table below.
RATING IN ART
Above average Below average Total
′
𝑨 𝑨
Above average
RATING 53 29 82
𝑴
IN
MATHEMATICS Below average
26 102 128
𝑴′
Total 79 131 210
A student is selected at random from these students.
a. What is the probability that the student will have:
1) a rating above average in art?
𝟕𝟗
𝑷(𝑨) = ≈ 𝟎. 𝟑𝟕𝟔𝟐
𝟐𝟏𝟎
2) a rating below average in mathematics?
𝟏𝟐𝟖
𝑷(𝑴′ ) = ≈ 𝟎. 𝟔𝟎𝟗𝟓
𝟐𝟏𝟎
b. What is the probability that the student will have:
1) a rating above average in mathematics given that the student has a rating above
average in art?
𝑷(𝑴∩𝑨) 𝟓𝟑⁄ 𝟓𝟑 53
𝑷(𝑴|𝑨) = = 𝟕𝟗 𝟐𝟏𝟎 = 𝟕𝟗 ≈ 𝟎. 𝟔𝟕𝟎𝟗 or from table: 𝑷(𝑴|𝑨) = 79 ≈ 𝟎. 𝟔𝟕𝟎𝟗
𝑷(𝑨) ⁄𝟐𝟏𝟎
2) a rating above average in art given that the student has a rating above average in
mathematics?
𝑷(𝑨∩𝑴) 𝟓𝟑⁄ 𝟓𝟑 𝟓𝟑
𝑷(𝑨|𝑴) = = 𝟖𝟐 𝟐𝟏𝟎 = 𝟖𝟐 ≈ 𝟎. 𝟔𝟒𝟔𝟑 or from table: 𝑷(𝑨|𝑴) = 𝟖𝟐 ≈ 𝟎. 𝟔𝟒𝟔𝟑
𝑷(𝑴) ⁄𝟐𝟏𝟎
c. What is the probability that the student will have:
1) a rating above average in both art and mathematics?
𝟓𝟑
𝑷(𝑨 ∩ 𝑴) = ≈ 𝟎. 𝟐𝟓𝟐𝟒
𝟐𝟏𝟎
2) a rating above average in art and a rating below average in mathematics?
𝟐𝟔
𝑷(𝑨 ∩ 𝑴′ ) = ≈ 𝟎. 𝟏𝟐𝟑𝟖
𝟐𝟏𝟎
1