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kanwar08
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 04 Oct 2024, 23:52
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KarishmaB I went through your post on "When Does Order Matter?". It's clearly mentioned that in case the the groups are different (men vs women, defective vs non-defective), in those cases we don't need to consider two different scenarios of who comes first.

Can you please clarify this once?


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Order matters because we can obtain 1 defective and 1 non defective in 2 ways. DN and ND.

When you pick 2 together from the same group of non defective fuses, say, and you say it can be done in 4/7*3/6 ways, you have already accounted for the possibility that either fuse could be picked first because 4 includes all non-defective fuses. So either could be picked first N1 or N2 (if you pick N1 and N2). The probability involves both cases.


But when you are picking from two different groups, one from defective and one from non defective, you have not accounted for both cases.

It is better to stick to combinations approach in these questions because 'when does order matter' is a tricky thing.
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 05 Oct 2024, 00:18
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And that is why I said it is tricky because every little thing will make a big difference. So I am unable to take care of every eventuality and give an exact set of rules. I have found that it ends up just confusing people and hence I have removed that post from my current blog.

When you pick a husband and a wife, you go to husband group to pick one and the wife group to pick one and which group you go to doesn't matter. They are not mixed up.
But when you are picking 2 fuses, you are doing it from the same group of 7 fuses but you are looking for the probability that the first was picked from the defective sub-group and second from non-defective sub group within that group of 7.
Hence the two cases are different.

If you want to explore this topic further, you can check various cases.
Calculate the probability of picking both defective, both non-defective and one defective one non-defective without counting this twice and notice that the probability doesn't add up to 1.
In other contexts it will add up to 1.
So you can develop an understanding of when it works.


kanwar08
KarishmaB I went through your post on "When Does Order Matter?". It's clearly mentioned that in case the the groups are different (men vs women, defective vs non-defective), in those cases we don't need to consider two different scenarios of who comes first.

Can you please clarify this once?


Quote:



Order matters because we can obtain 1 defective and 1 non defective in 2 ways. DN and ND.

When you pick 2 together from the same group of non defective fuses, say, and you say it can be done in 4/7*3/6 ways, you have already accounted for the possibility that either fuse could be picked first because 4 includes all non-defective fuses. So either could be picked first N1 or N2 (if you pick N1 and N2). The probability involves both cases.


But when you are picking from two different groups, one from defective and one from non defective, you have not accounted for both cases.

It is better to stick to combinations approach in these questions because 'when does order matter' is a tricky thing.
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 12 Nov 2024, 12:09
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How much time should one ideally take to comprehend and solve a question like this? Also do we expect the GMAT to supplement us with a question like this and solve it under two minutes?
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 13 Nov 2024, 01:08
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How much time should one ideally take to comprehend and solve a question like this? Also do we expect the GMAT to supplement us with a question like this and solve it under two minutes?
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The calculations involved in the question are not hard. They won't take much time. Yes, comprehension of options may take a little bit more. Overall, 2.5 - 3 mins max for any Quant question. After that one needs to bookmark and go ahead. Since the algorithm is adaptive, if one sees such a question, they are likely doing well on the test. It means they would have likely saved time on easy/medium questions and would have an extra minute to put in this. If it doesn't get solved within 2.5 - 3 mins then one needs to move on and come back to the question only if time permits. Even if one fails to solve a question like that, it won't impact their score much since the question is above their current ability. The issue arises when one invests 5 mins in a hard question (as per their current ability) and ends us missing out on easy/medium questions later. That is when the score would be below expectations.

To sum it all up, invest 2 - 2.5 mins per question. If you have extra time available because you were able to solve some previous questions fairly quickly, investing up to 3 mins is fine too. After that, bookmark, move on and come back if time permits.
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 04 Dec 2024, 10:32
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Total = 7, Defective = 3, Not defective = 4, To be chosen 2 without replacement

Probability of selecting 0 defective fuse = P(Z) = 4C2/7C2 = 4x3/7x6 = 4/14 = 2/7
Probability of selecting 1 defective fuse = P(O) = 3C1*4C1/7C2 = 4X3X2/7X6 = 4/7
Probability of selecting 2 defective fuse = P(T) = 3C2/7C2 = 3X2/7X6 = 1/7

The correct ratio of length should be Left : Middle : Right = 2 : 4 : 1.
The bar graph shows the ratio of length as Left : Middle : Right = 2 : 2 : 1.

We are told that 'The correct height is k times the height shown', thus the correct height is twice as high as the height shown, making k = 2, which is the middle bar.
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 21 Sep 2025, 21:10
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Why are we assuming the ball is drawn twice for all 3 cases?

MaxG00

Middle column:
P(1 def) = P(def & Ndef) + P(Ndef & def) = 3/7 × 4/6 + 4/7 × 3/6 = 4/7
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 21 Sep 2025, 23:58
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nandini14
Why are we assuming the ball is drawn twice for all 3 cases?


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This is a hard problem, and you should not be practicing hard problems if your fundamentals are weak.

First, the stem clearly says that two fuses are selected without replacement.

Second, I think your real confusion is not about the two draws but about why the probability for exactly 1 defective has two parts. The reason is that “1 defective and 1 non-defective” can happen in two different ways: the first draw is defective and the second is non-defective, or the first draw is non-defective and the second is defective. These are two distinct cases, and both must be counted.
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A box contains exactly 7 fuses, 3 of which are defective and 4 of 22 Sep 2025, 00:02
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nandini14
Why are we assuming the ball is drawn twice for all 3 cases?


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nandini14

Looking at your doubt, I can see there's a fundamental misconception about what's happening in this problem.

Key Clarification: We're NOT Drawing Twice for All 3 Cases

You're confusing "drawing 2 fuses" with "drawing twice for different cases." Here's what's actually happening:

The Actual Scenario:
  1. We reach into the box ONCE and pull out 2 fuses simultaneously (or one after another without replacement)
  2. This is ONE experiment with ONE drawing action of 2 fuses
  3. The three bars represent different possible outcomes of this SAME single experiment

Think of it This Way:
When you select \(2\) fuses from the box, you can get:

  • Outcome 1: Both non-defective (\(0\) defective) - Left bar
  • Outcome 2: One defective, one non-defective (\(1\) defective) - Middle bar
  • Outcome 3: Both defective (\(2\) defective) - Right bar
These are mutually exclusive outcomes of the same action of selecting \(2\) fuses.

Why the Calculations Look Like "Drawing Twice":

When we calculate \(P(\text{0 defective}) = \frac{4}{7} \times \frac{3}{6}\), we're calculating the probability of:

  • First fuse being non-defective: \(\frac{4}{7}\)
  • AND second fuse being non-defective: \(\frac{3}{6}\) (given the first was non-defective)

This multiplication represents the sequential nature of selecting \(2\) fuses, but it's still ONE drawing event that results in \(2\) fuses being selected.

Verification That These Are All From One Experiment:

Notice that: \(P(0 \text{ def}) + P(1 \text{ def}) + P(2 \text{ def}) = \frac{2}{7} + \frac{4}{7} + \frac{1}{7} = 1\)

The probabilities add up to \(1\) because these are all possible outcomes of the same experiment!

GMAT Strategy Tip: In probability problems involving multiple objects selected, always distinguish between:

The action (selecting \(n\) items)
The possible outcomes (different combinations you might get)

The bar graph shows the probabilities of different outcomes from the same action of selecting \(2\) fuses!
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