A certain club has 10 members, including Harry. One of the

Question

A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer. What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?

(A) 1/720
(B) 1/80
(C) 1/10
(D) 1/9
(E) 1/5

(A) 1/720
(B) 1/80
(C) 1/10
(D) 1/9
(E) 1/5

IanStewart · Answer

There are two different methods in that quote (I've numbered them in red). I wouldn't recommend using either for this type of problem -- it's easier just to think about making selections in order and multiplying, when possible. But you can use the above methods for the problem at the start of this thread:

1.: we have 10C3 possible choices in total for the first three people. If Harry will be one of our selections, we have 1C1 ways to choose him and 9C2 ways to choose the other two people, so the answer is 1C1*9C2/10C3 = 3/10

2.: we have 9C3 ways to choose 3 people different from Harry, and 10C3 selections possible in total, so the probability we don't pick Harry is 9C3/10C3 = 7/10, and the probability we do pick Harry is thus 1 - 7/10 = 3/10

1. : we have 10C3 possible choices in total for the first three people. If Harry will be one of our selections, we have 1C1 ways to choose him and 9C2 ways to choose the other two people, so the answer is 1C1*9C2/10C3 = 3/10

2. : we have 9C3 ways to choose 3 people different from Harry, and 10C3 selections possible in total, so the probability we don't pick Harry is 9C3/10C3 = 7/10, and the probability we do pick Harry is thus 1 - 7/10 = 3/10

IanStewart · Answer

Thanks to a PM from varotkorn, I realize I was answering the problem asked in the linked thread (but using the numbers in this one) rather than the question asked here (I got confused with so many tabs open linking to different questions!). I'll leave the above post up, but edit it to indicate what question I was answering. If instead you're answering the question asked in this thread, then you can still use the same methods, but we only care about the second and third selections, not the first, so the numbers get smaller. We have

1. : 10C2 ways to pick the Secretary and Treasurer, and 9C1*1C1 to make one of those selections Harry, so 9C1/10C2 = 1/5 is the answer

2. : 9C2 ways to pick the Secretary and Treasurer if Harry will not be selected, and 10C2 selections possible in total, so 1 - (9C2/10C2) = 1 - 4/5 = 1/5 is the answer

kornn · Answer

Dear  IanStewart 
 
I don't understand why we can completely ignore "president" selection?
How can we calculate as if there were no "president" selection at all?

IanStewart · Answer

The fact that they're picking a President doesn't matter, since we don't care if Harry is President. You can just imagine picking the Secretary and Treasurer first. Then it doesn't matter what they do with the remaining eight people - whether they give them all positions like Chairperson, President, Vice-President, etc, or give none of them positions. Harry will be just as likely either way to be Secretary or Treasurer.

This problem is identical, mathematically, to this one: you have 10 people, including Harry. They all line up. What is the probability Harry is 2nd or 3rd in line? Whether you call the first person in line "President" or don't call that person anything, the answer is the same either way.

mcelroytutoring · Answer

Some of you are overthinking this explanation!

As have others earlier on this thread, I recommend that you keep it simple. Don't count possibilities and make the math complicated--go straight to the probabilities, since that's the question being asked.

Concept: "or" probabilities can be added, so long as you remember to subtract "both" in the cases where that's an option.

Explanation: 1/10 + 1/10 = 2/10 = 1/5 
 
That's it.

-Brian

tkorzhan1995 · Answer

Bunuel ,  GMATNinja ,  NYCgirl15 , why we don't multiply by 7/8 to find

NOT President but Secretary = 9/10*1/9=1/10? (My understanding is that when selected as secretary, he cannot be selected as President or Treasurer. Negative prob for president is 9/10 and negative probability for treasurer is 7/8. Therefore, NOT President but Secretary = 9/10*1/9=1/10*7/8) Where I am getting wrong with my analysis?

President= 1/10
Not President= 1-1/10=9/10

Secretary= 1/9
Not Secretary= 1-1/9=8/9

Treasurer= 1/8
Not Treasurer= 1-1/8= 7/8

Thus,
1. NOT President but Secretary = 9/10*1/9=1/10
2. NOT President and NOT Secretary but Treasurer= 9/10*8/9*1/8=1/10
3. Either Secretary or Treasurer= 1/10+1/10=2/10=1/5

GMATNinja · Answer

By using "1/9" in your calculation of "NOT President but Secretary," you are  picking   Harry  as the Secretary. Because you've already  selected  him as Secretary, he is no longer part of the remaining 8 people.

In other words, none of the remaining 8 people are  Harry , so ANY of those 8 can be chosen as Treasurer (not just 7 of the 8).

I hope that helps a bit!

M838TE · Answer

IanStewart 
could you or someone help explain your approach on 1- P(not harry). I am not fully how 9C2 would show the various outcomes of picking not harry for treasurer or secretary. Are we just assuming since 1 of the 10 is picked for president (including harry as a possibility), then you subtract that 1 person from the 10, giving you 9 remnatns that could be treasurer or secretary? (what if Harry was among the 9)? please help thanks!

while the first approach somewhat made a better case for me that 1C1 for harry as a secretary or treasurer, since harry has been picked, the remaining 9 people could be chosen for the whichever position harry was not picked for ie., so you have a 9C1*1C1 way of arranging the desired outcome.
I get the concept of subtracting the chances that harry is not either of the position.
I understand the very simplified explanation but I thought the combinatoric formula helped me understand some underlying understanding of the stem.

IanStewart · Answer

Oh, that certainly wouldn't be "my approach" to this problem, as I hope I made clear. I said I "wouldn't recommend" this kind of approach to this problem, but perhaps I should have expressed that differently. Using combinatorics formulas here turns a simple problem into a needlessly complicated one.

I was only discussing that approach because kornn directed a question to me about why the method they were using didn't seem to work, so it was their approach that I was talking about, and I would never consider using those combinatorics formulas to answer a question like this. The answer to this question is instantly 2/10 if you look at it in the right way. If you think about this question:

• 10 people, including Harry, line up in a random order. What is the probability Harry is 3rd in line?

then the answer is 1/10, because Harry is equally likely to be in any position. If you then think of this question:

• 10 people, including Harry, line up in a random order. What is the probability Harry is 2nd or 3rd in line?

then the answer is 2/10, because in 2 of the 10 spots we can put Harry, he's 2nd or 3rd in line. If you then think about this question:

• 10 people, including Harry, line up in a random order, and the person 2nd in line will be called the "Treasurer" and the person third in line will be called the "Secretary". What is the probability Harry will be Treasurer or Secretary?

then I haven't changed the problem at all, so the answer is still 2/10.

All of that said, if you did want to understand why the answer to this question is also equal to 1 - (9C2 / 10C2) (and to reiterate, I'd never consider solving the problem this way), then it's important to understand what "9C2" and "10C2" mean. 10C2 means "the number of groups of two you can pick from a group of 10", where we don't care about order. So from the 10 people, there are 10C2 pairs of people we could choose to fill the secretary and treasurer positions, if we don't care which person is in which role. If we know we do not want Harry to occupy either of those roles, then we have fewer ways to fill them: we then need to fill the 2 roles from only 9 people. So we then have 9C2 pairs of people we could choose to fill the secretary and treasurer roles (again if order does not matter). So when we pick just two people to fill those two roles, the probability is 9C2/10C2 that Harry is  not  selected, and therefore 1 - 9C2/10C2 is the probability Harry  is  selected. But that's a very complicated way to solve the problem.

DaudDastagir · Answer

I am not strong in probability formula. However, I have an option to look at the problem from a plain arithmetic point of view. First of all the president will be chosen. If Harry hasn’t been chosen as the president, then he has the chance to be eihter secretary or the treasurer. Given that, once someone other than Harry has already been chosen as the president, now 9 members are remaining from whom first the secretary will be chosen. Similarly when the secretary has already been chosen, then the treasurer will be chosen from the remaining 8 members. Therefore, Harry's chance to be the secretary is 1/9, thus his chance of being the treasurer is 1/8. From this logic, the average of 1/9 and 1/8 is 17/72. 
I would really appreciate your help.

IanStewart · Answer

You can use probability rules to answer this question, though there are faster ways. For Harry to be chosen Secretary, two things need to happen: he must  not  be chosen President, and then he must be chosen Secretary. When we need a sequence of things to happen, we multiply the probabilities of each thing. Harry will not be chosen President 9/10 of the time, and then will be chosen Secretary 1/9 of the time, as you correctly found. Multiplying, the probability he is Secretary is 9/10 * 1/9 = 1/10. Similarly, to be chosen Treasurer, he must not be President, and must not be Secretary, then must be chosen Treasurer, so the probability he is Treasurer is 9/10 * 8/9 * 1/8 = 1/10. Then to find the probability he is either Secretary or Treasurer, we do not average those two probabilities; we add them (as long as they can't both happen). So the answer is 2/10.

As I said earlier in the thread, I think there are better ways to think about the problem. Harry is just as likely as anyone else to be chosen Secretary, regardless of whether they choose Secretary first or second or last, so there must be a 1/10 chance that happens, and the same is true of the Treasurer position.

DanTheGMATMan · Answer

Harry Potter!

https://www.youtube.com/watch?v=CBt_YvUkYU4v

ndwz · Answer

One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen at random to be the secretary, and one of the remaining 8 members is to be chosen at random to be the treasurer.

Hi - I got confused by this sentence because I thought this implied that the secretary and treasurer will be chosen in a trickle down order -- so 10 left, then 9 then 8 - what am i missing here

?

Bunuel · Answer

Yes, that’s how they are chosen: first 10, then 9, then 8. However, the procedure itself doesn’t affect the final calculation. There’s a detailed discussion of this question; I recommend reviewing it carefully. If something is still unclear, feel free to ask specific questions!

hjharsh · Answer

Hi Bunuel, I have understood the solution and have also gone through other practice questions that you recommended.

Wanted to understand the nCr approach here:

I am stuck with that- just to clarify the fundamentals:

( 9c1 * 1c1 * 8c8 + 9c1 * 9c9 * 1c1 ) / 10c3 --- I am getting that 6! from the denominator.

Can you explain the right approach?

Bunuel · Answer

Denominator should be 10P3 = 720, not 10C3.

Favorable cases:

Harry as secretary: 9*8 = 72 
 Harry as treasurer: 9*8 = 72 
 Total favorable = 144

Probability = 144/720 = 1/5.

egmat · Answer

This one is a great example of sequential selection with a hidden constraint that trips up many students.

Here's the key insight you need to see:  For Harry to become secretary or treasurer, he must first  avoid  being selected as president. If he's chosen as president, game over – he can't hold the other positions.

Let's break this down step by step:

Step 1: What's the probability Harry is NOT chosen as president?

When they pick the president, there are 10 people to choose from. Harry is 1 of them. So the chance Harry avoids the presidency is:

\frac{9}{10}

Think about it: 9 out of 10 people who aren't Harry could be president instead.

Step 2: Given Harry avoided presidency, what's his chance of getting secretary OR treasurer?

Now there are 9 people remaining (including Harry), and they need to fill 2 positions: secretary and treasurer.

Here's where students often stumble – you might think it's  \frac{1}{9} , but that's not quite right. Notice that there are  2 favorable positions  for Harry out of the 9 remaining people. So Harry's probability of being selected for one of these two roles is:

\frac{2}{9}

Step 3: Combine the probabilities

The overall probability is:

P(	ext{Harry is secretary or treasurer}) = P(	ext{not president}) 	imes P(	ext{secretary or treasurer | not president})

= \frac{9}{10} 	imes \frac{2}{9}

= \frac{18}{90}

= \frac{1}{5}

Notice how beautifully the 9's cancel out!

Answer: (E)  \frac{1}{5}

This makes intuitive sense too – Harry has a 20% chance, which feels reasonable given that 2 out of the 10 members will fill roles he wants, but he first needs to dodge that presidency selection.

Why this approach works:  You're dealing with conditional probability in a sequential selection process. The key is recognizing the constraint (Harry can't be president) and then calculating his chances within the remaining pool.

If you want to understand the systematic framework for tackling all sequential probability problems like this, including the common variations and time-saving patterns, you can check out the  complete solution breakdown on Neuron by e-GMAT . The full explanation covers alternative approaches and helps you recognize similar problem patterns instantly. You can also practice with detailed solutions for  other official GMAT questions here  to build consistent accuracy across all probability question types.

Hope this helps clarify the logic! Let me know if you have questions about any step.

bhanu29 · Answer

9 others + H

P(Harry Secretary) = (9C1 * 1 * 8C1) /(10*9*8) = 1/10
P(Harry Treasurer) = (9C1 * 8C1 * 1) / (10*9*8) = 1/10

Probability (Harry Secretary or Harry Treasurer) = 1/10 + 1/10 = 1/5

Correct Answer: E

luisdicampo · Answer

Deconstructing the Question

There are  10 members , including Harry. Three different positions are assigned in order: president, secretary, and treasurer.

We want the probability that Harry is chosen as either the secretary or the treasurer.

These are mutually exclusive cases, so we can add their probabilities.

Step-by-step

First, find the probability that Harry is chosen as secretary.

Harry must not be chosen as president, and then must be chosen as secretary:

\frac{9}{10} 	imes \frac{1}{9} = \frac{1}{10}

Now find the probability that Harry is chosen as treasurer.

Harry must not be chosen as president, must not be chosen as secretary, and then must be chosen as treasurer:

\frac{9}{10} 	imes \frac{8}{9} 	imes \frac{1}{8} = \frac{1}{10}

Add the two mutually exclusive cases:

\frac{1}{10} + \frac{1}{10} = \frac{2}{10} = \frac{1}{5}

Answer E

ischiragkapoor · Answer

Bunuel

I have solved a couple of question where they add different arrangements also but not in this one. I feel I know the reason but not sure. Can you help?

Like P(secretary)= P(NHN) = P(not Harry)*P(Harry)*P(not Harry)* 3!/2!
but you haven't considered the arrangements here, is it because the arrangement is given from prior, that first President, then Secretary and then the treasurer will be selected?

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