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# The events A and B are independent, and the probability that event A

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Re: The events A and B are independent, and the probability that event A [#permalink]
Official Explanation

We're asked to find the probability that event B occurs, such that one of the answer choices reflects the probability of either A or B occurring—in other words, any outcome other than neither A nor B occurring.

Since the only outcome we want to exclude is neither A nor B occurring (see stimulus explanation, above), we can create an equation by subtracting that one undesired outcome from the total of 1:

Probability that at least one of A and B occurs = 1 – [0.6(1 – b)]

= 1 – (0.6 – 0.6b)
= 1 – 0.6 + 0.6b
= 0.4 + 0.6b

(Note that there are two other ways to set this equation up. We could add the three desired outcomes together to yield 0.4b + 0.4(1 – b) + 0.6b. Alternately, we could add the probabilities of A and B, then subtract the double-counted outcome of both occurring, yielding 0.4 + b – 0.4b. On Test Day, go with whichever approach feels most straightforward—all of these approaches simplify to 0.4 + 0.6b.)

Now let's consider the possible values given in the answer choice list. Thinking about the situation logically, we can determine that the probability that B occurs must be less than the probability that at least one of A or B occurs. We will therefore begin by plugging in the smaller values from our list for the variable b in the equation we set up above. If one of these values for b results in a probability for "at least one of the events A or B" that matches one of the given choices, we know we have found the correct answer.

If b were equal to 0.1, then the probability that at least one of A and B occurs would equal 0.46, which is not an answer choice. This strongly suggests that 0.25 is the probability that B occurs, since it is the only other small answer choice. A larger probability might be possible, since the probability of at least one of A and B might be as high as 0.8, but plugging in b = 0.25 is a reasonable next step.

We'll substitute b = 0.25 into our equation and see whether it produces one of the other answer choices. If it does, that pair will be correct. If it doesn't, we'll try a different value for b. Plugging b = 0.25 into our equation yields:

Probability that at least one of A and B occurs = 0.4 + 0.6(0.25)

= 0.4 + 0.15
= 0.55

This possible value matches one of the answer choices, so we have found the values that satisfy the conditions. Now, as always with Two-Part Analysis questions, we must be careful to select the correct answer choices in the appropriate columns; it would be a shame to do all the math correctly but lose the points for this question by mixing up the columns.

The correct response for the "probability that at least one of the events A and B occurs" is 0.55; the correct response for the "probability that event B occurs" is 0.25.
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Re: The events A and B are independent, and the probability that event A [#permalink]
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Independent events are just two independent events..the existence on one happening does not affect that of the other...

So A happening= 0.4
B happening= x

I column:
At least any 1 should occur: So we see 'atleast one' and we go for the usual is 1- none

So 1- { P(Not A) * P(Not B) } => 1 - 0.6*P(Not B)
II column:
P(B) is what we need to find here

So we have two constraints one unknown
Start picking values such that both are satisfied..P(B) = 0.25 and P(atleast A or B) is 0.55..Am I correct here..

Quite a toughie..
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Re: The events A and B are independent, and the probability that event A [#permalink]
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What values are you picking? Still not clear on how you find P(B). Thanks!
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Re: The events A and B are independent, and the probability that event A [#permalink]
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FTS185 wrote:
What values are you picking? Still not clear on how you find P(B). Thanks!

Pick values one by one from the columns...for P(B)S: tart picking values from the right column and check whether a corresponding value for statement -I 1- [P(A)P(B)] is available in the left column..
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Re: The events A and B are independent, and the probability that event A [#permalink]
EMPOWERgmatRichC wrote:
We are told that they're independent events though - that's important since it effects how the math has to be done.

Hi EMPOWERgmatRichC,
Can you tell me what depedency we will have if they had specified that the events are dependent.
Still dont know differentiate between the two
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Re: The events A and B are independent, and the probability that event A [#permalink]
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Hi Shrivathsan,

'Dependent' variables in these types of probability questions are exceptionally rare on Test Day, so it's likely that you won't ever face that exact situation on Test Day.

Here's an example of that concept based on the probabilities in this question:

If the probability of Event A is 0.4 and probability of Event B is 0.25 (but ONLY if Event A happens first - otherwise, the probably of Event B is 0). What is the probability that at least one of the events occurs?

If the events were INDEPENDENT, then there would be 3 'satisfactory results' to consider:
1) A occurs, B does not
2) B occurs, A does not
3) Both A and B occur

However, since we have a dependent condition, there are only 2 'satisfactory results':
1) A occurs, B does not
2) Both A and B occur

There is NO situation in which 'B occurs, A does not', so we would have to adjust our math accordingly.

GMAT assassins aren't born, they're made,
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Re: The events A and B are independent, and the probability that event A [#permalink]
Thank you so much EMPOWERgmatRichC .
Now i understand the difference.

EMPOWERgmatRichC wrote:
Hi Shrivathsan,

'Dependent' variables in these types of probability questions are exceptionally rare on Test Day, so it's likely that you won't ever face that exact situation on Test Day.

Here's an example of that concept based on the probabilities in this question:

If the probability of Event A is 0.4 and probability of Event B is 0.25 (but ONLY if Event A happens first - otherwise, the probably of Event B is 0). What is the probability that at least one of the events occurs?

If the events were INDEPENDENT, then there would be 3 'satisfactory results' to consider:
1) A occurs, B does not
2) B occurs, A does not
3) Both A and B occur

However, since we have a dependent condition, there are only 2 'satisfactory results':
1) A occurs, B does not
2) Both A and B occur

There is NO situation in which 'B occurs, A does not', so we would have to adjust our math accordingly.

GMAT assassins aren't born, they're made,
Rich
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Re: The events A and B are independent, and the probability that event A [#permalink]
EMPOWERgmatRichC wrote:
Hi All,

For this question, we're given the probability that A occurs (0.4), but we're not given given the probability that B occurs. We are told that they're independent events though - that's important since it effects how the math has to be done.

Since we need to know the value of B to calculate the probability that EITHER A OR B OR BOTH will occur, we can solve this problem by TESTing THE ANSWERS. We're going to work our way down the list of values in the second column until we find a number that leads to one of the answers in the first column.

The easiest way to calculate the probability of "A or B or Both" is to actually calculate 1 - (not A)(not B).

So, starting in the second column...
If....
B = 0.10
1 - (not A)(not B) = 1 - (.6)(.9) = .46
That option does NOT exist in the first column

B = 0.25
1 - (not A)(not B) = 1 - (.6)(.75) = .55
That option DOES EXIST in the first column
This IS the correct set of answers.

0.55; 0.25

GMAT assassins aren't born, they're made,
Rich

Hi Rich ,
Thanks for the solution .
Just to better grasp the concept I was wondering how to arrive at the answer " The direct way ".

Probability of At least one event can be given by P(A) + P(B) +P(A)P(B) ..... either A or B or BOTH.
However when substituting the values I am getting 0.4+0.25 + 0.4*0.25= 0.75
what exactly am I forgetting here?
should I be subtracting P(A)P(B).

Very much appreciate your help thank you.
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Re: The events A and B are independent, and the probability that event A [#permalink]
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Hi stne,

If you're going to be literal about calculating what is asked for, then you have to make some changes to the equation that you created. We're looking for 3 things:

1) The probability that A occurs, but B does NOT
2) The probability that B occurs, but A does NOT
3) The probability that BOTH A and B occur

IF....
Probability of A = .4
Probability of B = .25

Then the total probability of the three possibilities listed above is...

(.4)(1 - .25) + (.25)(1 - .4) + (.4)(.25) =
(.4)(.75) + (.25)(.6) + (.4)(.25) =
.300 + .150 + .100 =
.55

GMAT assassins aren't born, they're made,
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Re: The events A and B are independent, and the probability that event A [#permalink]
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Edit: I ended up figuring it out, Kaplan just solved it in an odd way I didn't know.

P(a or b) = 0.4 +b - 0.4b
P(a or b) = 0.4 + 0.6b

Then just plug in the values until you find out that 0.25=b and .55=P(a or b)

Does anyone have suggestions on how to do this question? This question is from Kaplan and their solution is quite confusing:
We're asked to find the probability that event B occurs, such that one of the answer choices reflects the probability of either A or B occurring—in other words, any outcome other than neither A nor B occurring.

Since the only outcome we want to exclude is neither A nor B occurring (see stimulus explanation, above), we can create an equation by subtracting that one undesired outcome from the total of 1:

Probability that at least one of A and B occurs = 1 – [0.6(1 – b)]

= 1 – (0.6 – 0.6b)

= 1 – 0.6 + 0.6b

= 0.4 + 0.6b

(Note that there are two other ways to set this equation up. We could add the three desired outcomes together to yield 0.4b + 0.4(1 – b) + 0.6b. Alternately, we could add the probabilities of A and B, then subtract the double-counted outcome of both occurring, yielding 0.4 + b – 0.4b. On Test Day, go with whichever approach feels most straightforward—all of these approaches simplify to 0.4 + 0.6b.)

Now let's consider the possible values given in the answer choice list. Thinking about the situation logically, we can determine that the probability that B occurs must be less than the probability that at least one of A or B occurs. We will therefore begin by plugging in the smaller values from our list for the variable b in the equation we set up above. If one of these values for b results in a probability for "at least one of the events A or B" that matches one of the given choices, we know we have found the correct answer.

If b were equal to 0.1, then the probability that at least one of A and B occurs would equal 0.46, which is not an answer choice. This strongly suggests that 0.25 is the probability that B occurs, since it is the only other small answer choice. A larger probability might be possible, since the probability of at least one of A and B might be as high as 0.8, but plugging in b = 0.25 is a reasonable next step.

We'll substitute b = 0.25 into our equation and see whether it produces one of the other answer choices. If it does, that pair will be correct. If it doesn't, we'll try a different value for b. Plugging b = 0.25 into our equation yields:

Probability that at least one of A and B occurs = 0.4 + 0.6(0.25)

= 0.4 + 0.15

= 0.55

This possible value matches one of the answer choices, so we have found the values that satisfy the conditions. Now, as always with Two-Part Analysis questions, we must be careful to select the correct answer choices in the appropriate columns; it would be a shame to do all the math correctly but lose the points for this question by mixing up the columns.

The correct response for the "probability that at least one of the events A and B occurs" is 0.55; the correct response for the "probability that event B occurs" is 0.25.
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Re: The events A and B are independent, and the probability that event A [#permalink]
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The events A and B are independent, and the probability that event A occurs is 0.4.

In the table below, choose the two numbers that are consistent with the information that is given. In the first column, select the row that shows the probability that at least one of the events A and B occurs, and in the second column, select the row that shows the probability that event B occurs.

 Probability that at least one of the events A and B occurs Probability that event B occurs 0.10 0.25 0.50 0.55 0.60 0.80

Probability that at least one of the events A and B occurs = 0.55
Probability that event B occurs = 0.25

Originally posted by padwals on 26 Jun 2020, 23:50.
Last edited by padwals on 21 Aug 2020, 22:06, edited 3 times in total.
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Re: The events A and B are independent, and the probability that event A [#permalink]
Why can't we just use P(A or B) = P(A) + P(B) here since events are independent?
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Re: The events A and B are independent, and the probability that event A [#permalink]
Monkeyy wrote:
Why can't we just use P(A or B) = P(A) + P(B) here since events are independent?

P(A or B) = P(A) + P(B) includes repetition of one case where A happens as well as B happens therefore we need to exclude that case,
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Re: The events A and B are independent, and the probability that event A [#permalink]
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Since both A and B are independent events
And probability of A = P(A)=0.4

So probability of atleast one of event A and B occur= X = 1-(1-P(A))(1-P(B)) i.e 1- both doesn't occur

Now we will put the values of P(B) and X from the given options ,so the option fitting the answer choice is
X=.55
P(B)=.25

Posted from my mobile device
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Re: The events A and B are independent, and the probability that event A [#permalink]
GMATinsight wrote:
Monkeyy wrote:
Why can't we just use P(A or B) = P(A) + P(B) here since events are independent?

P(A or B) = P(A) + P(B) includes repetition of one case where A happens as well as B happens therefore we need to exclude that case,

Can you please explain it in little detail, as i am still confused
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Re: The events A and B are independent, and the probability that event A [#permalink]
I would like to give my 2 cents here. Now this is a little high level topic that i am goin to discuss. (Probably Bunuel will understand )

No offence, but the below APPROCH (not solution) of EMPOWERgmatRichC maybe wrong. Because we are not given that A & B are complimentary events. In simple terms we are not given that only 2 events A & B are present in the universe. In that case we should apply formulae P(A or B) = P(A) + P(B) - P(A and B).

Not P(A or B) = 1 - (not A)(not B) [What if there is 3rd element C is present!!!! ]

Though the answer might come same with both the approaches in this case which indicates that 2 events are complimentary(i.e only 2 events A & B are present in the universe ). But since it is not given in the question we must not assume this!!!

EMPOWERgmatRichC wrote:
Hi All,

For this question, we're given the probability that A occurs (0.4), but we're not given given the probability that B occurs. We are told that they're independent events though - that's important since it effects how the math has to be done.

Since we need to know the value of B to calculate the probability that EITHER A OR B OR BOTH will occur, we can solve this problem by TESTing THE ANSWERS. We're going to work our way down the list of values in the second column until we find a number that leads to one of the answers in the first column.

The easiest way to calculate the probability of "A or B or Both" is to actually calculate 1 - (not A)(not B).

So, starting in the second column...
If....
B = 0.10
1 - (not A)(not B) = 1 - (.6)(.9) = .46
That option does NOT exist in the first column

B = 0.25
1 - (not A)(not B) = 1 - (.6)(.75) = .55
That option DOES EXIST in the first column
This IS the correct set of answers.

0.55; 0.25

GMAT assassins aren't born, they're made,
Rich
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Re: The events A and B are independent, and the probability that event A [#permalink]
KarishmaB wrote:
stne wrote:
EMPOWERgmatRichC wrote:
Hi All,

For this question, we're given the probability that A occurs (0.4), but we're not given given the probability that B occurs. We are told that they're independent events though - that's important since it effects how the math has to be done.

Since we need to know the value of B to calculate the probability that EITHER A OR B OR BOTH will occur, we can solve this problem by TESTing THE ANSWERS. We're going to work our way down the list of values in the second column until we find a number that leads to one of the answers in the first column.

The easiest way to calculate the probability of "A or B or Both" is to actually calculate 1 - (not A)(not B).

So, starting in the second column...
If....
B = 0.10
1 - (not A)(not B) = 1 - (.6)(.9) = .46
That option does NOT exist in the first column

B = 0.25
1 - (not A)(not B) = 1 - (.6)(.75) = .55
That option DOES EXIST in the first column
This IS the correct set of answers.

0.55; 0.25

GMAT assassins aren't born, they're made,
Rich

Hi Rich ,
Thanks for the solution .
Just to better grasp the concept I was wondering how to arrive at the answer " The direct way ".

Probability of At least one event can be given by P(A) + P(B) +P(A)P(B) ..... either A or B or BOTH.
However when substituting the values I am getting 0.4+0.25 + 0.4*0.25= 0.75
what exactly am I forgetting here?
should I be subtracting P(A)P(B).

Very much appreciate your help thank you.

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

Think about why you subtract P(A and B). The concept is parallel to the sets concept. The Probability that both occur is counted twice - once in P(A) and then in P(B). So you subtract it out once to ensure no double counting.

Since A and B are independent events, P(A and B) = P(A) * P(B)

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

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

P(A or B) = 0.4 + P(B) - 0.4*P(B)

P(A or B) = 0.4 + 0.6*P(B)

Now, look at what can be the value of P(A or B). It will be something more than 0.4. So try 0.5 for it.
doesn't work because 0.5 = 0.4 + 0.6*P(B) gives P(B) = .167 which is not one of the options for P(B).

Try 0.55 for P(A or B). You get P(B) = 0.25. It is there in the options.

So P(B) = 0.25 and P(A or B) = 0.55 is one of the many possible solutions.

KarishmaB Bunuel

In general in a probability question, how would we know that P(A) and P(B) can occur together or not? Or whether P(A intersection B) is 0 or not? So if a question is P(A) or P(B), so when should we subtract P(A and B) and when not? Is that clearly given in question that there is intersection ?
Re: The events A and B are independent, and the probability that event A [#permalink]
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