The events A and B are independent. The probability that event A occu

The events A and B are independent. The probability that event A occurs is 0.6, and the probability that at least one of the events A or B occurs is 0.94. What is the probability that event B occurs?

(A) 0.34
(B) 0.65
(C) 0.72
(D) 0.76
(E) 0.85

As given independent event p( A and B) = pA *pB
p(AorB) = 0.94


p(AorB) = pA + pB - p( A and B)
= pA + pB - pA *pB
0.94 =0.6 + pB - (0.6*pB)

On solving the above we get
pB = 0.85
E ans .

Probability of A
- Yes: 0.6
- No: 0.4

Probability of B
- Yes: Y
- No: N

Probability of at least A or B = 1 - (0.4)(N)
0.94 = 1 - (0.4)N
N = 0.15

1 - N = Y
1 - 0.15 = Y
Y = 0.85

Answer Choice E

KAPLAN OFFICIAL SOLUTION:

In order to find the probability that event B occurs in this problem, we need to set up an equation that includes the probabilities we are given and which allows us to solve for B. We are told that the probability that at least one of A or B occurring is 0.94. ‘At least one of A or B’ means that an outcome is desired if A occurs and B does not, B occurs and A does not or A and B both occur.

It is important to remember two rules of probability here. First, when you encounter an ‘or’ situation you add and when you see an ‘and’ situation you multiply. Second, the probability that an event does NOT occur is equal to one minus the probability it does occur.

Based on these rules, we can translate ‘at least one of A or B occurs is 0.94’ to the following equation:

.6B [the probability that both A and B occur] + .6(1-B) [the probability that A occurs and B does not] + B(1-.6) [the probability that B occurs and A does not] = .94

We can simplify this equation to .6B + .6(1-B) + .4B = .94 and solve for B.

.6B + .6(1-B) + .4B = .94

.6B + .6 – .6B + .4B = .94

.6 + .4B = .94

.4B = .34

B = .34/.4

B = 34/40 = .85

Thus, our answer is 0.85, or answer choice (E).

very very tricky one indeed.
P(A)=0.6
P(A or B) = 0.94
since both are independent, P(A and B) is P(A)*P(B)
using OR formula we get:
P(A or B) = P(A) + P(B) - P(A and B)
0.94 = 0.6 + P(B) - P(B)*0.6
0.34 = P(B) - 0.6*P(B)
factor P(B)
0.34 = P(B)(1-0.6)
0.34 = P(B)*(0.4)
PB = 0.34/0.4
use fractions better
34/40 -> 17/20 - multiply by 5 -> 85/100 so 0.85 - E

Hi!
I have an alternative solution.

Prob A won't occur =1-0.6=0.4
Let's assume that Prob B won't occur x.

Then, Prob both A and B together will not occur is 0.4x=1-0.94=0.06
So, x=0.15

Thus, Prob B will occur is 1-0.15=0.85

Hence, E.

P(at least one of them) = P(A) + P(B) - P(A x B)
0.94 = 0.6 + B - 0.6xB
0.94 = 0.6 + 0.4B
0.34 = 0.4B
B = 0.34/0.4 (same as 17/20) = 0.85 E.

We can use the formula:

P(A or B) = P(A) + P(B) - P(A and B)

Since A and B are independent, P(A and B) = P(A) * P(B). Therefore, we have:

P(A or B) = P(A) + P(B) - P(A) * P(B)

0.94 = 0.6 + P(B) - 0.6 * P(B)

0.34 = 0.4 * P(B)

P(B) = 0.34/0.4 = 0.85

Answer: E

This the way I solved and went wrong:

Only A = A
Only B = B
Intersection = C
None = D

A+B+C = 0.94

I assumed B+C is what is asked as P(B). However, B+C = 0.34.
While I understand the correct approach, can someone explain why this leads to wrong result?

Very nice High Quality question

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