In a research study, each of 15 participants has been given exactly

Hi MartyMurray
Can you please explain the solution for this.
Thanks in advance

Thanks for the explanation! Can you please help me understand how did you arrive at the approximate value of 0.9^8? Because as per my understanding exam onscreen calculator does not have the facility to calculate powers. Is there any alternate way for it?

Two approaches:

- Use the GMAT calculator and just keep multiplying your result by 0.9 until you have multiplied 0.9 by itself 8 times.

- Do the math in your head and approximate the results using the 9 multiplication table values:

.9 x .9 = .81

.8 x .9 = .72

.7 x .9 = .63

.6 x .9 = .54

.5 x .9 = .45

.4 x .9 = .36

.3 x .9 = .27

Notice that we rounded down a few times and still the final result, .27, is greater than .20. So, we can be confident that the exact result would have been greater than .2.

That rather rough rounding method happens to work in this case. Probably the question was set up so that approximation would work that way with rounding, but if you're handling a similar question, you may be better off using the calculator 7 times in case the approximated value isn't clearly greater or less than the value you're comparing it to. Alternatively, you could be more exact with the math than I was and not just round down every time.

Hi MartyMurray , Thanks for the explanation.

Can you tell me why can't we do like this for Participant 6:
Attempts: 8
Probability of producing a correct image in an attempt: 0.10
Probability of producing a correct image in all attempts: 0.1x8 = 0.8

That I was assuming 80% probability of getting correct in 8 attempts.
Hence satisfying the hypothesis.

Please let me know where I am being ignorant here.
Thanks!

Thank you so much for the explanation!

It appears that in a research study, 15 participants were assigned 15 printing devices, each intended to print a specific image. However, these devices frequently malfunction and fail to produce the intended image. Each participant, armed with data on their device's past performance, has chosen a certain number of attempts they believe will provide a good chance of producing at least one correct image. The table likely includes each participant's chosen number of attempts and the actual probability of their device producing a correct image on any single attempt.

Hi AvikSaha.

Here's one way to see the issue.

The probability of getting heads when we flip a coin is 0.50. So, if we flip a coin twice, does the probability of getting heads become 100%? No, right? In fact, even if we flip the coin 5 times, the probability of getting at least 1 heads is not 100%. If you think about it, you would not bet your house that, if you flip a coin 5 times, 1 time you will get heads. You will likely get 1 heads, but the probability is not 100%. Also, it's certainly not 5 × 0.50 = 250%.

So, we can see that the probability of an event occurring at least once in a given number of attempts cannot be the probability that that event will occur multiplied by the number of attempts.

So, what's going on?

What's going on is that what we really have with multiple attempts is multiple sets of possible outcomes.

For example, if we flip a coin twice, we have these 4 possible sets of outcomes of our 2 coin flips:

HH HT TT TH

Since there's a heads in 3 of those sets, we can see that the probability of getting at least 1 heads in 2 flips is 75%. There's also 1 set that includes no heads. So, even though the probability of getting a heads in 1 flip is 50%, the probability of getting a heads in 2 flips is not 100%.

What occurs when the probability of printing correctly is 0.10 and we print 8 times is similar. It's not the case that 80% of the sets of possible outcomes of 8 attempts include a correctly printed image. Rather, only around 57% of the sets of possible outcomes of 8 attempts include at least one correctly printed image.

So, the probability that we will get at least one correctly printed image in 8 attempts is around 57%.

Thank you so much MartyMurray for such a detailed explanation. It make sense to me now.

Please one more querry on this question.
Why are we doing like this "Probability of not producing a correct image in all attempts: (0.90)^8" ?
instead of directly doing Probability of producing a correct image in all attempts: (0.10)^8"

To calculate the probability of producing at least 1 correct image, we'd have to consider not just the probability of producing a correct image in all 8 attempts, which is (0.10)^8, as you said. That's the probability of producing 8 correct images, which is not the probability we want.

Rather we'd have to consider the probabilities of producing 1, 2, 3, 4, 5, 6, 7, and 8 images in 8 attempts, since in each of those cases, we'd have at least 1 correct image.

It's much simpler to calculate the probability of producing no correct images and then subtract that probability from 1.

Can we calculate the probability of success (for example in the first case) by elevating 0.1 to 8?

MartyMurray

That approach won't work, and if you think about it, you can see why.

The higher the power we raise 0.1 to, the lower the result. So, if we use that approach, the more attempts we had, the lower would be the probability of success, an outcome that doesn't make sense.

In a research study, each of 15 participants has been given exactly 1 of 15 printing devices. All 15 devices are intended to print a certain image, but each frequently malfunctions and fails to produce the intended image. Each participant has examined data about the past performance of his or her device and has chosen a number of attempts that he or she believes will be the minimum needed to have a good chance of producing at least one correct image. For each participant, the table shows the number of attempts that the participant will make and the actual probability of that participant’s device producing a correct image on any single attempt.

Hypothesis: Every participant has at least an 80 percent probability of producing the correct image within the number of attempts that he or she has chosen.

For each of the following participants, select Contradicts the hypothesis if the data shown in the table for that participant, considered alone, contradicts the hypothesis. Otherwise, select Does not contradict the hypothesis.

Paticipant 6

Participant 7

Participant 9


bb if you can correct the formatting. Each Participant has two options. Contradicts the Hypothesis or Doesnot Contradict the Hypothesis

Hello Experts,

Please find below my approach .

Participant 6

Probability of success =0.1 . So Probability of failure =0.9

Hence , probability of success in 8 attempts = (0.9 ^7) *0.1 =  0.047

Similarly for other participants.

But , OG solution is different. 
Please guide .

The reason you are missing out so many other solutions.
One success in 8 attempts: 0.048 but this could be in any of the 8 attempts. So, P1 = 8C1 * 0.048 = 0.384
Two successes in 8 attempts: \((0.9)^6*(0.1)^2\) but this could be in any two of the 8 attempts. So, \(P2 = 8C2 * (0.9)^6  * (0.1)^2 = 0.148\)
Next is around 0.03 and will become lesser and lesser.
Total = \(0.382+0.148+0.03 = 0.56\) and very small amounts will get added to it to come to \(0.57\)

I hope it settles your query.

The question is already discussed here
https://gmatclub.com/forum/in-a-research-study-each-of-15-participants-has-been-given-exactly-424099.html

Simple probability rules and some calculator work:

Need to know:

for independent events P(Y) + P(N) = 1

Also review theory of mutually exhaustive events if you don't already know it. It helps with both probability and combinatorics.

Participant 6: P(Success) = 0.1 P(Failure) = 0.9

What's the probability of failing 8 attempts in a row? (fail AND fail AND fail AND fail ...)

0.9 * 0.9 * 0.9 .... 8 times on a calculator (notated as 0.9^8)

0.9^8 = 0.43. There is a 43% chance that participant 6 will fail 9 times.

1 - 0.43 = .57 or 57% chance the participant will succeed AT LEAST ONCE during the 8 attempts. This doesn't fall within at least 80% chance. Mark as contradicts.

Participant 7

0.5 * 0.5 = 25% chance of failure on two attempts

1-.25 = 0.75 or 75% chance of success on at least one attempt. Contradicts.

Participant 9

1 - 0.26 = 0.74

0.74^3 = 0.405

1-0.405 = .595 or 59.5% chance of success. Not > 80% chance of success. Contradicts.­­

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