A box contains exactly 7 fuses, 3 of which are defective and 4 of

I solved this problem logically.

There are 4 non-defective fuses and 3 defective fuses. So, if we draw 1 fuse the probability of drawing a non-defective fuse will be higher than that of drawing a defective fuse. Similarly, because of this asymmetrical nature (4 vs 3) the probability of drawing 2 non-defective fuses (0 defects: left column) will be higher than the probability of drawing 2 defective fuses (2 defects: right column). So, neither C nor E cannot be the answer (as in that case left and right columns will have equal height). That leaves us with A, B and D.
Now the probability of drawing a mixed collection (1 defective + 1 non-defective) will be higher than drawing a 'pure' collection (2 defective or 2 non-defectives). So, middle column will have higher elevation than the rest. That means neither A nor D can be the answer as doing so will be the probability of drawing 2 non-defective fuses is higher than drawing 1 defective fuse and 1 non-defective fuse.
B is the answer.

The middle bar's probability in the graph represents the probability of 2/7. But Actually the probability would be 4/7 because there are two different ways to arrange. Either they pull out a defective fuse first then a non defective fuse OR vice versa they pull out a non defective fuse followed by a defective fuse. Also note, when you hear the word "OR" in probability, that means you add instead of multiply.

Therefore the middle bar should be doubled.­

Middle column:
P(1 def) = P(def & Ndef) + P(Ndef & def) = 3/7 × 4/6 + 4/7 × 3/6 = 4/7  
 

Given: A box contains exactly 7 fuses, 3 of which are defective and 4 of which are not defective. In the bar graph above, the left and middle bars are the same height and the right bar is half this height. The heights are intended to represent, respectively, the probabilities of obtaining 0 defective fuses, exactly 1 defective fuse, and 2 defective fuses when 2 fuses are randomly selected without replacement from the box. However, 1 of the bar heights is incorrect. The correct height is k times the height shown.

Asked: Which of the following choices for the bar and the value of k would result in an accurate representation of the labeled probabilities?

The probabilty of obtaining 0 defective fuses when 2 fuses are randomly selected without replacement from the box = 4/7*3/6 = 2/7
The probabilty of obtaining 1 defective fuses when 2 fuses are randomly selected without replacement from the box = 3/7*4/6 + 4/7*3/6 = 4/7
The probabilty of obtaining 2 defective fuses when 2 fuses are randomly selected without replacement from the box = 3/7*2/6 = 1/7

Middle bar, k = 2

IMO B
­

­Your explanation is very helpful, I just want to clear out some points.

1. In the calculation of the probabilities, only the second one has two scenarios and its second term (4/7 × 3/6) is calculated irrespective of the first one? (that after taking one defective the new total is 6 I mean, because in this point when I solved it I thought that since there isn't replacement, the second term I mentioned earlier should be 2/4 for the defective and 3/5 for the non defective)

2. In the last row, if we multiply the middle bar by 2 then the ratios will be 2:8:1 which is incorrect. Since the first two must be equal we should multiply the middle bar by k=1/2 so as to be 2:2:1

Your points above seem inconsistent with each other, but it's unclear since you don't complete the probability calculation for your first point.

In any event, the two ways 1 defective and 1 nondefective item can be drawn should be thought of as one way OR the other.

So that when the defective is drawn first and the nondefective second:

3/7 * 4/6

the number of items available to be drawn from is reduced by 1 after the defective item is drawn.

This scenario is distinct from drawing the non defective item first and the defective item second:

4/7 * 3/6

So, drawings from the first scenario do not reduce the pool for the second, since they are alternative ways of achieving the objective.

So the total probability is the sum of the two scenarios:

4/7

Since the actual probability for selecting 0 defectives is 2/7 and for 1 defective 4/7, the bar in the middle needs to be doubled.

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For my first question that's what I meant.

For the second query, selecting 0 defectives is the left bar with prob = 2/7 and selecting 1 is the middle with prob = 4/7.

Since it has to be equal, if we double the middle one then they'll be 2/7 for the left column and 8/7 for the middle. So we have to divide by 2 (or in other words multiply by k=1/2) the middle one so as to be both 2/7 which satisfies the condition that the right one (1/7) is half of the other two.

­

Ok I see.

You have truth and the representation of truth (chart) mixed up.

Truth is 2/7, 4/7 and 1/7, as borne out by the calculations.

The chart is attempting to represent the above.

What about the chart is incorrect and what needs to be done ?

The first and second bars are the same height, indicating the same probability for each, not correct.

We can halve the first bar so that the relationship is corrected.

But then the 1st and 3rd bars would be the same height, suggesting equal probability, which we know isn't correct.

So, if we double the 2nd bar instead, it will indicate twice the probability of the first and four times that of the third, all of which correctly represents the true probabilities.

Posted from my mobile device

­Hang on a sec, let's take a step back.

We know from the stem that the left and middle bars have to be equal in length (so equal probabilities) and at the same time the right one must be exactly half of the other two. We are also aware that there is a mistake in one of these lengths and we try to spot and correct it. Lastly, we know that the left bar depicts the probability to select 0 defectives, the middle one the probability to choose 1 defective and the right the probability to pick 2 defectives. We calculated these probabilities and we see that:

Prob (0 defect) = 2/7 = p1
Prob (1 defect) = 4/7 = p2
Prob (2 defect) = 1/7 = p3

In order to satisfy the conditions I mentioned earlier it must be true that:

(1) p1 = p2 => 2/7 = 4/7  AND  (2) p3 = 1/2*p1 = 1/2*p2

We can conclude that the first condition can be satisfied if we multiply the p1 by k = 2 or multiply the p2 by k=1/2.

If we choose the first one, then the second condition is incorrect since it becomes 1/7 = 1/2*(2*2/7) = 1/2*4/7 = 2/7 , whereas if we choose the second one then the condition No2 is valid since it becomes 1/7 = (1/2)*(1/2)*(4/7) = 1/7.

So, why is it wrong to multiply the middle bar by 1/2 since, as it seems above, it provides the results we want contrary to those when we multiply the middle bar by 2?

No. The first and second bars do not have to be the same length.

The question stem says "intended".

The objective is to make the bars represent reality, the actual probabilities.

Posted from my mobile device

­Exactly and the stem says that the height is the probability and it also mentions "In the bar graph above, the left and middle bars are the same height and the right bar is half this height". So that's what we have to represent, am I right?

You have to make the bars represent 2/7,4/7 and 1/7, that is, ratios of 2,4 and 1.

The heights currently shown are irrelevant, except that there is a factor to be applied to one of them k.

The problem could have shown no bars and the objective would be the same: plot the bar heights whose relative heights represent the relative probabilities.

­I am missing the point also why we need to multiply by 2 and not by 1/2. can anyone elaborate a bit more on the reasoning?

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.

Hope this helps.