A random study of recent graduates' average grades and degrees showed the following results.
Grade
Degree C B A
B.S. 6 9 13
B.A. 5 12 10
If a graduate is selected at random, find these probabilities.
1.The graduate has a B.S. degree, given that he or she has an A average.
2.Given that the graduate has a B.A. degree, the graduate has a C average.
3.What is the probability that a person has a B.S. degree and a B?
4.What is the probability that a graduate has neither an A nor a C.
Bs 7/48
12
To find the probabilities for the given questions, we will use conditional probability. Conditional probability is the probability of an event occurring, given that another event has already occurred. We will also use the concept of total probability.
Before we proceed, let's find the total number of graduates in the study to use it in our calculations. We add up all the frequencies to get the total count:
Total Graduates = Sum of all frequencies = 6 + 9 + 13 + 5 + 12 + 10 = 55
Let's now solve each question:
1. The graduate has a B.S. degree, given that he or she has an A average:
P(B.S. and A) = Number of B.S. graduates with A average / Total Graduates
= 13 / 55
= 0.2364 (approximately)
2. Given that the graduate has a B.A. degree, the graduate has a C average:
P(C | B.A.) = Number of B.A. graduates with C average / Total B.A. graduates
= 5 / (5 + 12 + 10)
= 5 / 27
≈ 0.1852 (approximately)
3. Probability that a person has a B.S. degree and a B:
P(B.S. and B) = Number of B.S. graduates with B grade / Total Graduates
= 6 / 55
= 0.1091 (approximately)
4. Probability that a graduate has neither an A nor a C:
P(neither A nor C) = 1 - P(A) - P(C)
P(A) = Number of graduates with A average / Total Graduates
= (13 + 10) / 55
= 23 / 55
P(C) = Number of graduates with C average / Total Graduates
= (6 + 5) / 55
= 11 / 55
Substituting the values:
P(neither A nor C) = 1 - (23 / 55) - (11 / 55)
= (55 - 23 - 11) / 55
= 21 / 55
≈ 0.3818 (approximately)
These are the probabilities as per the given data.