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

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Math Expert
Joined: 02 Sep 2009
Posts: 64130
The events A and B are independent. The probability that event A occu  [#permalink]

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11 Sep 2015, 00:59
3
23
00:00

Difficulty:

75% (hard)

Question Stats:

55% (02:12) correct 45% (01:35) wrong based on 221 sessions

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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

Kudos for a correct solution.

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Re: The events A and B are independent. The probability that event A occu  [#permalink]

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11 Sep 2015, 01:55
5
P(A U B) = P(A) + P(B) - P(A intersection B)

for independent events, the probability of both occuring

P(A intersection B) = P(A) * P(B)

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

So, P(B) = 0.85

##### General Discussion
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Joined: 02 Sep 2015
Posts: 4
Re: The events A and B are independent. The probability that event A occu  [#permalink]

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11 Sep 2015, 01:47
3
1
At least one of the events A or B occurs means
1) Event A occur (Ao)and Event B does not (Bd)
2) Event B occur (Bo) and Event A does not (Ad)
3) Both events A and B occur (Ao & Bo)

Hence 0.94 = Ao*Bd + Bo*Ad + Ao*Bo

Does not occur is equal to one minus the probability it occurs

WKT The probability that event A occurs is 0.6 ie Ao = 0.6

0.94= 0.6*(1-Bo)+ Bo*(1-Ao) + Ao*Bo
0.94=0.6*(1-Bo) + Bo*(1-0.6)+ 0.6*Bo
On simplification we get,
The probability that event B occurs (Bo) = 0.85
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Re: The events A and B are independent. The probability that event A occu  [#permalink]

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11 Sep 2015, 12:32
2
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 .
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The events A and B are independent. The probability that event A occu  [#permalink]

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12 Sep 2015, 13:48
1
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

Math Expert
Joined: 02 Sep 2009
Posts: 64130
Re: The events A and B are independent. The probability that event A occu  [#permalink]

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14 Sep 2015, 05:03
1
1
Bunuel wrote:
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

Kudos for a correct solution.

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

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Re: The events A and B are independent. The probability that event A occu  [#permalink]

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27 Mar 2016, 19:39
1
Bunuel wrote:
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

Kudos for a correct solution.

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
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Re: The events A and B are independent. The probability that event A occu  [#permalink]

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11 Apr 2018, 03:53
Bunuel wrote:
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

Kudos for a correct solution.

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.
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Re: The events A and B are independent. The probability that event A occu  [#permalink]

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22 Jan 2020, 07:43
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.
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Re: The events A and B are independent. The probability that event A occu  [#permalink]

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24 Jan 2020, 09:23
Bunuel wrote:
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

Kudos for a correct solution.

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

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Posts: 28
Re: The events A and B are independent. The probability that event A occu  [#permalink]

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28 Mar 2020, 01:06
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?
Re: The events A and B are independent. The probability that event A occu   [#permalink] 28 Mar 2020, 01:06