Explanation

Since you know that the probability of event A is \(\frac{1}{2}\)and the probability of event B is \(\frac{1}{3}\) but you are not given any information about the relationship between events A and B, you can compute only the minimum possible value and the maximum possible value of the probability of the event A ∪ B. The probability of A ∪ B is least if B is a subset of A; in that case, the probability of A ∪ B is just the probability of A, or \(\frac{1}{2}\) .

The probability of A ∪ B is greatest if A and B do not intersect at all; in that case, the probability of A ∪ B is the sum of the probabilities of A and B, or \(\frac{1}{2} + \frac{1}{3} = \frac{5}{6}\) With no further information about A and B, the probability that A or B, or both, will occur could be any number from \(\frac{1}{2}\) to \(\frac{5}{6}\).
Of the answer choices given, only \(\frac{1}{2}\) and \(\frac{3}{4}\) are in this interval.

The correct answer consists of Choices B and C.
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Dear Carcass, please could you help me…

Since we don’t know the relationship between A and B my assumption is based on the following

Two extreme cases:

1. If the events A and B are not mutually exclusive, then p(A or B) = p(A) + p(B) - p(A and B).
p(A or B) = ½+1/3-1/6=2/3

2. If events A and B are mutually exclusive, then the probability of A or B is simply:
p(A or B) = p(A) + p(B).
p(A or B) = ½+1/3=5/6
Thus, I assumed correct answer should be just C

This is the correct expression

P(A ∪ B) = P(A) + P(B) - P(B)P( A | B )

P(A) is probability of A happening
P(B) is probability of B happening
P( B | A ) can be said to Probability of B given that A has occured. Now look at the figure below. The area P( A | B ) is minimum when P(A) and P(B) do not overlap at all.

.
Intutive explanation of the formula:

P(A ∪ B) = P(A) + P(B) - P(B)P( A | B )

If you look at the figure to find area of P(A ∪ B) we add the area of P(A) and P(B) but if P(A) and P(B) overlap, we would add the overlapping area twice! So we sunbtract P(B)P( A | B ).

Example:

An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn without replacement from the urn. What is the probability that both of the marbles are black?


Let A = the event that the first marble is black; and let B = the event that the second marble is black.

We know the following:

In the beginning, there are 10 marbles in the urn, 4 of which are black. Therefore, P(A) = 4/10.
After the first selection, there are 9 marbles in the urn, 3 of which are black. Therefore, P(B|A) = 3/9.

P(A) P(B|A) = (4/10) * (3/9) = 12/90 = 2/15 = 0.133


P(B)P( A | B ) is not P(B)P(A). The amount of overlapping area cannot be found out unless specified

Please take care of this as this is the very fundamental concept of probability. Hope this helps and feel free to ask more questions.
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It is a confusing question.

The probability of event A occurring without event B occurring is P(A) * (1-P(B) = 1/2 * (1-1/3) = 2/6
The probability of event B occurring without event A occurring is P(B) * (1-P(A) = 1/3 * (1-1/2) = 1/6
The probability of both event A & B occurring is P(A) * P(B) = 1/2 * 1/3 = 1/6

Thus, The probability of event A ∪ B (that is, the event A or B, or both) occurring is the sum of the above 2/6+1/6+1/6 = 5/6.

No answer is given!

The answers are B and C.

It is an official question.

Regards
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I agree with the official explanation, but there was a similar question https://greprepclub.com/forum/probabili ... -8733.html wherein we had a discussion about the minimum probability of the events.
Why is the option 1st also not correct? Since the event B can occur with 1/3 probability

Probability of event \(B\) is \(0.5\)
Probability of event \(A\) is \(0.3333\)

Given that probability of event \(B\) is greater than \(A\)
one A U B scenario could involve event \(A\) being a subset of \(B\) hence minimum possible union will be \(0.5\)
However if the two probabilities are mutually exclusive \(A + B\) would be \(0.8333\) which would be the maximum probability
as such the range of probability for A U B is \(0.5\) to \(0.8333\) inclusive
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Event B has more terms than event A, shouldn't A be a subset of B, which gives minimum probability as 1/3?

-----------ASIDE-----------------
Key Concept #1: P(J OR K) = P(J) + P(K) - P(J and K)

Key Concept #2: P(J and K) ≤ P(J) & P(J and K) ≤ P(K)

This should make sense.
For example, P(it rains AND it snows tomorrow) must be less than or equal to P(it rains tomorrow)
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GIVEN:
P(A) = 1/2
P(B) = 1/3

From Key Concept #2 above, we know that P(A and B) ≤ 1/2, and we know that P(A and B) ≤ 1/3
So, the greatest possible value of P(A and B) is 1/3...
And the least possible value of P(A and B) is 0


We want to calculate P(A OR B)
P(A OR B) = P(A) + P(B) - P(A and B)
= 1/2 + 1/3 - P(A and B)

If P(A and B) = 1/3, then P(A OR B) = 1/2 + 1/3 - 1/3 = 1/2
If P(A and B) = 0, then P(A OR B) = 1/2 + 1/3 - 0 = 5/6

So, the RANGE of possible values of P(A OR B) goes from 1/2 to 5/6 (inclusive)

Check the answer choices....B and C are within the range of 1/2 to 5/6 (inclusive)

Answer: B & C

Cheers,
Brent
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-------> For a certain probability experiment, the probability that event A will occur is 1/2 and the probability that event B will occur is 1/3. Which of the following values could be the probability that the event A ∪ B (that is, the event A or B, or both) will occur?


THIS IS CONFUSING! SOMEONE HELP ME PLEASE

Don't worry about the "A ∪ B" notation. You aren't expected to know it on test day.
On test day, the question will as us to find P(A or B).

Also note that this is a VERY tricky question (to date, only 33% of students have correctly answered the question.

Cheers,
Brent
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Since we get 2/3 and 5/6, each equals 0.66 and 0.83 respectively, hence only option C comes between these 2 integers.

How you got \(\frac{2}{3}\)?
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Prob. (A will occur) = \(\frac{1}{2}\)
Prob. (A will not occur) = \(1 - (\frac{1}{2}) = \frac{1}{2}\)

Prob. (B will occur) = \(\frac{1}{3}\)
Prob. (B will not occur) = \(1 - (\frac{1}{3}) = \frac{2}{3}\)

1. Prob. (only A will occur) = P (A will occur) x P (B will not occur) = \((\frac{1}{2})(\frac{2}{3}) = \frac{1}{3}\)

2. Prob. (only B will occur) = P (B will occur) x P (A will not occur) = \((\frac{1}{3})(\frac{1}{2}) = \frac{1}{6}\)

3. Prob. (both A and B will occur) = P (A will occur) x P (B will occur) = \((\frac{1}{2})(\frac{1}{3}) = \frac{1}{6}\)

4. Prob. (either A or B will occur) = P (A will occur) x P (B will not occur) + P (B will occur) x P (A will not occur) = \((\frac{1}{3}) + (\frac{1}{6}) = \frac{1}{2}\)

Only 2 option match our cases.
Hence, option B and C
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