Question
At a certain university, 30% of the students major in math. Of the students majoring in math, 60% are males. Of all the students at the university, 70% are males.
a) What is the probability that a student selected at random in the university is male and majors in math?
b) What is the probability that a randomly selected student in the university is female or majors in math or both?
Based upon the information you provided, out of every 100 students you have this breakdown:
male, math major: 18
female, math major: 12
male, non-math major: 52
female, non-math major: 18
therefore:
a) 18%
b) female: 30%
math major (either sex): 30%
female math major: 12%
so for b), you just add them up to get 72%?
a) What is the probability that a student selected at random in the university is male and majors in math?
b) What is the probability that a randomly selected student in the university is female or majors in math or both?
Based upon the information you provided, out of every 100 students you have this breakdown:
male, math major: 18
female, math major: 12
male, non-math major: 52
female, non-math major: 18
therefore:
a) 18%
b) female: 30%
math major (either sex): 30%
female math major: 12%
so for b), you just add them up to get 72%?
Answers
this probably wrong, see correct answer below
P(Math) = 0.30
P(Male | Math) = 0.60
P(Male) = 0.70
P(Male ∩ Math) = P(Male | Math) ⋅ P(Math) = 0.60 ⋅ 0.30 = 0.18
P(Female) = P(¬Male) = 1 − P(Male) = 1 − 0.70 = 0.30
P(Math) = 0.30
P(Female | Math) = P(¬Male | Math) = 1 − P(Male | Math) = 1 − 0.60 = 0.40
P(Female ∩ Math) = P(Math) ⋅ P(Female | Math) = 0.30 ⋅ 0.40 = 0.12
P(Female ∪ Math) = P(Female) + P(Math) − P(Female ∩ Math) = 0.30 + 0.30 − 0.12 = 0.48
P(Math) = 0.30
P(Male | Math) = 0.60
P(Male) = 0.70
P(Male ∩ Math) = P(Male | Math) ⋅ P(Math) = 0.60 ⋅ 0.30 = 0.18
P(Female) = P(¬Male) = 1 − P(Male) = 1 − 0.70 = 0.30
P(Math) = 0.30
P(Female | Math) = P(¬Male | Math) = 1 − P(Male | Math) = 1 − 0.60 = 0.40
P(Female ∩ Math) = P(Math) ⋅ P(Female | Math) = 0.30 ⋅ 0.40 = 0.12
P(Female ∪ Math) = P(Female) + P(Math) − P(Female ∩ Math) = 0.30 + 0.30 − 0.12 = 0.48
A single, standard number cube is tossed. What is the probability of getting a 4 or 5?
A. 1
B. 1/5
C. 1/3
D. 1/6
A. 1
B. 1/5
C. 1/3
D. 1/6
The number cube has 6 equally likely possible outcomes, which are the numbers 1, 2, 3, 4, 5, and 6. Only two of these outcomes, 4 and 5, satisfy the condition of getting a 4 or 5. Therefore, the probability of getting a 4 or 5 is 2 out of 6, or 1/3.
Answer: C. 1/3
Answer: C. 1/3
A game involves spinning this spinner. What is the probability of the pointer landing on R?
A. 1/4
B. 1/2
C. 1/8
D. 3/8
A. 1/4
B. 1/2
C. 1/8
D. 3/8
Since the spinner has 8 equal sections, the probability of the pointer landing on any one of the sections is 1/8. Since the section labeled R covers only one of the 8 sections, the probability of the pointer landing on R is 1 out of 8, or 1/8.
Answer: C. 1/8
Answer: C. 1/8
The table shows the drink preferences of 50 shoppers at the mall. What is the probability that one shopper, selected at random from the 50 surveyed, preferred either Drink B or Drink A?
Drink. |. Number of shoppers
A. |. 7
B. | 8
C. |. 10
D. |. 11
E. |. 14
Drink. |. Number of shoppers
A. |. 7
B. | 8
C. |. 10
D. |. 11
E. |. 14
The total number of shoppers surveyed is 50. The number of shoppers who preferred Drink A or Drink B is the sum of the number of shoppers who preferred each drink: 7 + 8 = 15. Therefore, the probability of a shopper selected at random from the 50 surveyed preferring either Drink A or Drink B is 15 out of 50, or 3 out of 10, or 0.3.
Answer: Probability = 0.3
Answer: Probability = 0.3
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