I29-168

Question

At a wildlife sanctuary, researchers track the survival of two newly released animal groups. Group A has a 3/5 chance of surviving its first winter, while Group B has a 4/5 chance of surviving its first winter.
 
Select for  Minimum Survival  the lowest probability, consistent with these estimates, that both groups survive the winter. Select for  Maximum Survival  the highest probability, consistent with these estimates, that both groups survive the winter. Make only two selections, one in each column.

Bunuel · Answer

Official Solution:

• Minimum Survival Probability 
 
We use the equation:
 
Total = P(A) + P(B) – P(both) + P(neither).
 
Substitute the values:
 
1 = 3/5 + 4/5 – P(both) + P(neither).
 
P(both) = 2/5 + P(neither).
 
To minimize P(both), we set P(neither) = 0.
 
This gives P(both) = 2/5.
 
 Alternative we can do the following: 
 
We are given:
   
P(A) = 3/5 (the probability that Group A survives),
 P(B) = 4/5 (the probability that Group B survives).   
This means the probability that Group B does not survive is:
   
P(not B) = 1 - P(B) = 1/5.
   
Now consider the event “A survives  and  B does not survive.” Its probability cannot be greater than either P(A) = 3/5 or P(not B) = 1/5. Therefore, the  maximum  possible probability of this event, “A survives  and  B does not survive,” is 1/5.
 
But P(A) = 3/5 represents all cases where A survives, and that breaks into two disjoint parts:
   
A survives and B survives, and
 
A survives and B does not survive.
   
So we can write:
   
P(A) = P(A and B) + P(A and not B).
   
We know P(A) = 3/5, and the maximum possible value of P(A and not B) is 1/5.
 
Therefore, the minimum possible value of P(A and B) is:
   
P(A and B) = 3/5 - 1/5 = 2/5.
   
So by this reasoning as well, the minimum survival probability for both is 2/5.
 
 • Maximum Survival Probability 
 
The probability of Group A surviving is 3/5 (60%) and the probability of Group B surviving is 4/5 (80%). The maximum probability that both survive cannot be greater than either one individually. Therefore, the maximum possible value of P(both) is the smaller of the two: 3/5 (60%).

Correct answer:

Minimum Survival  " \frac{2}{5} "

Maximum Survival  " \frac{3}{5} "

Manvi01 · Answer

I like the solution - it’s helpful.

natusrerum · Answer

Concept for this question explained by Karishma B.: GMAT Quant - Solving Compound Probability Questions ( Search on Yt, unable to post links)

sunshineeee · Answer

I’d like to confirm the question choice in the table both asking maximum survival, should it the minimum and maximum survival instead? Thanks

Bunuel · Answer

I think it should be fixed now. Thank you!

sunshineeee · Answer

Yeay, thanks Bunuel!

SunnyRun · Answer

I like the solution - it’s helpful.

abhikc3004 · Answer

I like the solution - it’s helpful.

FATT · Answer

Ok i solved this using a game theory table

Step 1: Set up the table. Let x = P(both A and B survive). Given P(A) = 3/5 and P(B) = 4/5, use the marginals to build a 2×2 joint probability table:

Step 2: Express every cell in terms of x. Once one cell is set as x, the other three are forced by the marginals

Step 3: Apply the constraint — every cell ≥ 0. Probabilities can't be negative, so:

x ≥ 0
3/5 − x ≥ 0 → x ≤ 3/5
4/5 − x ≥ 0 → x ≤ 4/5
x − 2/5 ≥ 0 → x ≥ 2/5

Combining: 2/5 ≤ x ≤ 3/5

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zitipiti · Answer

I like the solution - it’s helpful.

Minimum Survival	Maximum Survival
		\(0\)
		\(\frac{1}{5}\)
		\(\frac{2}{5}\)
		\(\frac{1}{2}\)
		\(\frac{3}{5}\)
		\(\frac{4}{5}\)

Minimum Survival	Maximum Survival
		\(0\)
		\(\frac{1}{5}\)
		\(\frac{2}{5}\)
		\(\frac{1}{2}\)
		\(\frac{3}{5}\)
		\(\frac{4}{5}\)

I29-168

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