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# The events A and B are independent. The probability that both events A

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Intern
Joined: 02 Apr 2018
Posts: 20
The events A and B are independent. The probability that both events A  [#permalink]

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10 Jun 2018, 08:45
6
00:00

Difficulty:

55% (hard)

Question Stats:

59% (02:15) correct 41% (02:17) wrong based on 59 sessions

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The events A and B are independent. The probability that both events A and B occur is 0.21. The probability that event A occurs and event B does not occur is 0.49. What is the probability that at least one of the events A and B occur?

a) 0.58
b) 0.79
c) 0.84
d) 0.89
e) 0.93
ISB, NUS, NTU Moderator
Joined: 11 Aug 2016
Posts: 343
Re: The events A and B are independent. The probability that both events A  [#permalink]

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10 Jun 2018, 09:15
4
praneethuddagiri wrote:
The events A and B are independent. The probability that both events A and B occur is 0.21. The probability that event A occurs and event B does not occur is 0.49. What is the probability that at least one of the events A and B occur?

a) 0.58
b) 0.79
c) 0.84
d) 0.89
e) 0.93

To Calculate:
The probability that at least one of the events A and B occur.
Which will be equivalent to :
Probability that Event A occurs and B does not occurs + Probability that Event B occurs and A does not occurs + Probability that Event A and B both occur .
=A*(1-B) + B*(1-A) + A*B

The probability that event A occurs - A
The probability that event B occurs - B

Given:
The probability that both events A and B occur= 0.21 ==A*B........................................(i)
The probability that event A occurs and event B does not occur = 0.49 ==A*(1-B)...........(ii)

A*(1-B)=0.49
A-AB=0.49
A-0.21=0.49 Using (i)
Therefore, A=0.70
And Hence, B=0.30

Now, Calculating the probability that at least one of the events A and B occur
A*(1-B) + B*(1-A) + A*B
0.49 + 0.30*0.30 + 0.70+0.30
0.49 + 0.09 + 0.21
0.79

Alternatively, We can also solve by calculating the Probability that none of the Events A or B Occur
The probability that at least one of the events A and B occur = 1- Probability that none of the Events A or B Occur
1 - (1-A)*(1-B)
1 - (1-0.7)*(1-0.3)
1 - 0.3*0.7
1 - 0.21
0.79
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Re: The events A and B are independent. The probability that both events A  [#permalink]

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10 Jun 2018, 10:09
praneethuddagiri wrote:
The events A and B are independent. The probability that both events A and B occur is 0.21. The probability that event A occurs and event B does not occur is 0.49. What is the probability that at least one of the events A and B occur?

a) 0.58
b) 0.79
c) 0.84
d) 0.89
e) 0.93

Let's write all the given data in mathematical notation:-

P(A)P(B)=0.21----------(1)
P(A)(1-P(B))=0.49------(2)
Question stem:-
P(at least one of A & B occur)=1-P(neither A nor B occurs)

from(2), we have P(A)-P(A)P(B)=0.49
Or,P(A)-0.21=0.49
Or, P(A)=0.7

So P(B)=$$\frac{0.21}{0.7}$$=0.3

P(neither A)=1-P(A)=1-0.7=0.3
P(neither B)=1-P(B)=1-0.3=0.7

So, P(at least one of A & B occur)=1-P(neither A nor B occur)=1-0.3*0.7=1-0.21=0.79

Ans. B
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Intern
Joined: 17 Jun 2017
Posts: 18
Re: The events A and B are independent. The probability that both events A  [#permalink]

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26 Jun 2018, 02:42
1
We are given:
P(A and B) = .21
P(A and not B) = .49

Then, we know P(A) = P(A and B) + P(A and not B) = .7

because A and B are independent
P(A and B) =P(A)*P(B)= .21
Plug in .7 into P(A) we get P(A)*P(B)= .7*P(B) = .21 and we know P(B) = .3
Then P(not A and B) = P(B) - P(A and B) = .3 - .21 = .09

Now, probability of at least one event occur is P(A and not B) + P(not A and B) + P(A and B) = .49 + .09 + .21 =.79
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Joined: 04 Aug 2010
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Re: The events A and B are independent. The probability that both events A  [#permalink]

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26 Jun 2018, 05:43
2
praneethuddagiri wrote:
The events A and B are independent. The probability that both events A and B occur is 0.21. The probability that event A occurs and event B does not occur is 0.49. What is the probability that at least one of the events A and B occur?

a) 0.58
b) 0.79
c) 0.84
d) 0.89
e) 0.93

To organize the data, we can use the following Double-Matrix:

The probability that both events A and B occur is 0.21.
The probability that event A occurs and event B does not occur is 0.49

Since (both A and B) = 0.21, and A but not B) = 0.49, we get:

(both A and B) is equal to (A)(B):
(A)(B) = 0.21
(A but not B) is equal to (A)(1-B):
(A)(1-B) = 0.49.
If we divide the first equation by the second, we get:
$$\frac{B}{(1-B)} = \frac{0.21}{0.49}$$
$$\frac{B}{(1-B)} = \frac{21}{49}$$
$$\frac{B}{(1-B)} = \frac{3}{7}$$
$$7B = 3 - 3B$$

$$10B = 3$$

$$B = 0.3$$

If we insert (total B) = 0.3 into the matrix and calculate the remaining values, we get:

Since (neither A nor B) = 0.21, (at least A or B) = 1 - 0.21 = 0.79.

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Re: The events A and B are independent. The probability that both events A &nbs [#permalink] 26 Jun 2018, 05:43
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