There are 5 treatments for a disease: A, B, C, D and E. If a doctor ch

Question

There are 5 treatments for a disease: A, B, C, D and E. If a doctor chooses 3 out of 5 for the treatment, what is the probability probability at least one of A and B will be chosen?

A. 1/10
B. 1/5
C. 3/10
D. 3/5
E. 9/10

GMATinsight · Answer

Total possible random selection of 3 out of treatments = 5C3 = 10

The ways of choosing three treatments such that  none  of A and B is chosen = 3C3 (choose 3 out of CD and E) = 1

So unfavourable Probability such that neither A nor B is in selected treatments  = \frac{1}{10}

Favorable Probability such that atleast one of A and B is selected  = 1-(\frac{1}{10}) = (\frac{9}{10})

Bunuel  I think there is some mistake in options...Please check

Bunuel · Answer

_____________________
Typo edited. Thank you!

EgmatQuantExpert · Answer

Solution

Given  
In this question, we are given that
 • There are 5 possible treatments for a disease: A, B, C, D, and E
• A doctor chooses 3 out of these 5

To find  
We need to determine
 • The probability that at least one of A and B will be chosen

Approach and Working out

• At least one of A and B is chosen
 o If we directly consider this, there will be multiple possibilities such as - 
  Only A is selected
 Only B is selected
 Both A and B are selected

o Computing probability from this will take time. The best way is to find the probability of the opposite situation and then subtract it from 1.
o The opposite of “at least one of A and B is chosen” is “none of A and B is chosen”
  Only possibility: C, D, E
 • Probability of choosing {C, D, E} =  \frac{^3C_3}{^5C_3}  =  \frac{1}{10}  
 Hence, desired probability = 1 –  \frac{1}{10}  =  \frac{9}{10} .

Thus, option E is the correct answer.

Correct Answer: Option E

Pritishd · Answer

1. Probability = Favorable outcomes / Total outcomes
2. Now, probability of choosing A or B = 1 - (probability of choosing B, C and D / probability of choosing 3 out of 5 treatments)
3.  1 - (3C3 / 5C3) = 1 - (1/10) = 9/10

Ans.  E

ScottTargetTestPrep · Answer

Solution:
 
There are 5C3 = (5 x 4 x 3) / (3 x 2) = 10 ways to choose 3 treatments out of 5. Since the total number of ways is only 10, let’s list those that contain at least one of A and B:

ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BDE

As we can see from the above, the only treatment combo that doesn’t contain either A or B is CDE. Therefore, the probability a treatment will contain at least one of A and B is 9/10.

Alternate Solution:
 
There are 5C3 = (5 x 4 x 3) / (3 x 2) = 10 ways to choose 3 treatments out of 5. The only outcome of the 10 possible ways that doesn’t include either of A or B is (C, D, E), so the probability that either A or B is selected is 9/10.

Answer: E

There are 5 treatments for a disease: A, B, C, D and E. If a doctor ch

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