GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
The probability that event A occurs is 0.4, and the probabil
[#permalink]
01 Dec 2018, 08:27
01 Dec 2018, 08:27
Expert Reply
6
Bookmarks
00:00
Question Stats:
83% (02:11) correct
16% (01:41) wrong based on 137 sessions
Hide Show timer Statistics
The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: The probability that event A occurs is 0.4, and the probabil
[#permalink]
03 Dec 2018, 11:24
03 Dec 2018, 11:24
2
1
Bookmarks
Carcass wrote:
The probability that event A occurs is 0.4, and the probability that events A and B both occur is 0.25. If the probability that either event A or event B occurs is 0.6, what is the probability that event B will occur?
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
A. 0.05
B. 0.15
C. 0.45
D. 0.50
E. 0.55
Given:
P(A) = 0.4
P(A and B) = 0.25
P(A or B) = 0.6
Let's apply the OR probability formula: P(A or B) = P(A) + P(B) - P(A and B)
Plug values into the formula to get: 0.6 = 0.4 + P(B) - 0.25
Simplify to get: 0.6 = 0.15 + P(B)
Solve: P(B) = 0.45
Answer: C
Re: The probability that event A occurs is 0.4, and the probabil
[#permalink]
30 Apr 2020, 02:17
30 Apr 2020, 02:17
1
A small doubt here, it says the probability of EITHER A or B and not EITHER A or B or BOTH, so don't we have to subtract the intersection twice? Because we have to exclude the probability that both occur together completely.
If P(A) = 0.4 and P(A and B) is 0.25, then P(only A not B) is 0.4 - 0.25 = 0.15.
Now P(A or B and not both) is 0.6 = P(only A not B) + P(only B not A) = 0.15 + P(only B not A)
So P(only B not A) = 0.6 - 0.15 = 0.45.
Now it asks for P(B) so it includes both P(A and B) and P(only B not A) = 0.45 + 0.25 = 0.7.
It doesn't mention whether we want the probability of B with or without A here.
If P(A) = 0.4 and P(A and B) is 0.25, then P(only A not B) is 0.4 - 0.25 = 0.15.
Now P(A or B and not both) is 0.6 = P(only A not B) + P(only B not A) = 0.15 + P(only B not A)
So P(only B not A) = 0.6 - 0.15 = 0.45.
Now it asks for P(B) so it includes both P(A and B) and P(only B not A) = 0.45 + 0.25 = 0.7.
It doesn't mention whether we want the probability of B with or without A here.
