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DanTe02
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A randomly selected sample population consists of 60% women 08 Jan 2021, 23:17
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chetan2u
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1) take ways in which 1 out of 3 or 2 out of 3 or 3out of three are colorblind- a LONG method
a) 1 out of 3 - .6*.4*.4 *3 = .288
b) 2 out of 3 - .6*.6*.4*3 = .432
c) 3 out of 3 - .6*.6*.6 = .216
add all three .288+.432+.216 = .936
we have multiplied a and b by 3 because in three ways the colorblind or non-colorblind can be choosen - CNN, NCN and NNC
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I believe there's some room of debatable issue here. The question says scientist are picking till they get a colorblind subject. With this logic, Option A , B, And C make a mistake. In option A The probability you calculated shows A case where scientists picked a Non Color blind even after getting a color blind. same issue with B and C.
The same thing can be repeated for people doing 1-.4*.4*.4
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A randomly selected sample population consists of 60% women 08 Jan 2021, 23:27
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chetan2u
Quote:
1) take ways in which 1 out of 3 or 2 out of 3 or 3out of three are colorblind- a LONG method
a) 1 out of 3 - .6*.4*.4 *3 = .288
b) 2 out of 3 - .6*.6*.4*3 = .432
c) 3 out of 3 - .6*.6*.6 = .216
add all three .288+.432+.216 = .936
we have multiplied a and b by 3 because in three ways the colorblind or non-colorblind can be choosen - CNN, NCN and NNC
I believe there's some room of debatable issue here. The question says scientist are picking till they get a colorblind subject. With this logic, Option A , B, And C make a mistake. In option A The probability you calculated shows A case where scientists picked a Non Color blind even after getting a color blind. same issue with B and C.
The same thing can be repeated for people doing 1-.4*.4*.4
Show more


Yes, you are correct. May be when I posted the solution, I misread the question.
The ways will be
NNC-0.4*0.4*0.6=0.096
NC-0.4*0.6=0.24
C-0.6
Total 0.096+0.24+0.6=0.936
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A randomly selected sample population consists of 60% women 05 Feb 2025, 14:37
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Many seem confused by the replacement vs not replacement issue.

If the size of the population had been specified then it would make sense NOT to put that person back into the population for the next selections.

So if there were 60 and we're selecting 1 then that would leave 59 and the probability would be adjusted for that revised base.

Likewise if there were 600 then the base for the next probability would be revised to 599.

But the two probabilities would then not be the same and so the population size makes a difference when not replacing.

So knowing the population size is required when not replacing.

Therefore the only way to calculate the probabilities is to assume replacement because then our not knowing the population size doesn't matter.
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A randomly selected sample population consists of 60% women 07 Jan 2026, 02:23
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Hi Bunuel, how is this question different from this :

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Since I do not know the previous outcomes
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A randomly selected sample population consists of 60% women 07 Jan 2026, 03:16
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AditiDeokar
Hi Bunuel, how is this question different from this :

A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?

The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).

Since I do not know the previous outcomes
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Other than both questions involving probability, I do not think they have anything else in common.

In this question, what confuses many people is whether the selection is with replacement or without replacement. However, it does not matter here. First, the question asks for an approximate probability. Second, we are selecting from a population, and the implied assumption is that it is large enough that the with replacement versus without replacement distinction is not important, especially since the question asks for an approximate probability.

For example, suppose the population is 1,000 people and 40% (400 people) are not colorblind. On the first pick, the probability of selecting a not colorblind person is 400/1,000. If that person is removed, the probability on the next pick becomes 399/999, which is still approximately 40%. So, for this problem, the distinction does not really matter.
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A randomly selected sample population consists of 60% women 08 Jan 2026, 01:53
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Deconstructing the Question
Goal: Find the approximate probability of selecting a colorblind person in no more than 3 tries.

Step 1: Calculate the Probability of Selecting a Colorblind Person (P(C))
The population is a mix of Women (W) and Men (M).
* \(P(W) = 0.60\)
* \(P(M) = 0.40\)

Conditional probabilities of being colorblind:
* \(P(C | W) = 0.90\)
* \(P(C | M) = 0.15\)

Using the Weighted Average (Total Probability Rule):
\(P(C) = (0.60 \times 0.90) + (0.40 \times 0.15)\)
\(P(C) = 0.54 + 0.06\)
\(P(C) = 0.60\)

So, there is a 60% chance of success on any single try.
Conversely, there is a 40% chance of failure (picking a non-colorblind person, \(P(NC) = 0.40\)).

Step 2: Calculate Probability of Success in 3 Tries
It is faster to calculate the probability of the complement event: Failing 3 times in a row.
\(P(\text{Fail 3 times}) = 0.40 \times 0.40 \times 0.40 = 0.064\) (or 6.4%).

The probability of success (at least once in 3 tries) is:
\(P(\text{Success}) = 1 - P(\text{Fail 3 times})\)
\(P(\text{Success}) = 1 - 0.064 = 0.936\)

Step 3: Match with Options
\(0.936\) is \(93.6\%\).
Comparing to the choices:
(A) 95%
(B) 90%

93.6% is closer to 95% (diff 1.4) than to 90% (diff 3.6).

Answer: A
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