Home » 40 Questions on Probability for data science

# 40 Questions on Probability for data science

• Kenneth Singh says:

Hi,

None of the options in question 7 are probability values right? I think all of them are Numerators. Shouldn’t the correct option be “None of the above”?

• Chen Luo says:

Maybe this problem should be ‘the number of possible cases’.

• Aditya Royal Matturi says:

The answer for question 7 is (1/52 c 4)*(48 c 4/52 c 4) please correct that

• Dishashree Gupta says:

Yes, we have removed 7 from score calculation.

• Arcady Novosyolov says:

Hello,

Thanks for publishing this set of problems. It looks like the answer to the question 21 is right (option A), but solution is wrong, since actual probabilities of events are P(A) = 0.402, P(B) = 0.296, and P(C) = 0.245, see explanation at https://github.com/arcadynovosyolov/math_and_prob/blob/master/note_on_question_21.ipynb

Best regards,

• santiago says:

Why you do that? If you ask something with a real example please give a real answer. Do you really think that the probability is 0.33?
The result is 0.5. The events are independant.

• Dishashree Gupta says:

This is a problem of conditional probability and we have defined the case where one child is already a girl. Now the sample set reduces and hence we have only 3 choices in the sample set.

• santiago says:

I understand the idea, but if your objective is to test conditional probability maybe another example could be better.
In real life the sex of one of your childs doesn’t affect the sex of those who will come

• Aditya Royal Matturi says:

As we are concerned about statistics here it definitely effects statistically, I am also sure that it doesn’t effect genetically and it’s pure luck and as always luck as has some role in many things like tossing coins etc but we are ignorant of it.

• David Harper says:

Hi Dishashree, The flaw is that {BB, GG, BG, GB} represent a permutation. When you reveal that the first is a girl, you are revealing GX or XG and excluding BB and BX so that 2/4 remain. Put another way, there are 3 combinations {BB, BG or GB, and GG} such that revealing one girl eliminates BB and what remains is 1/2. I hope that’s helpful!

• Sam Gu says:

There are 2 variants of this question:
1. Alice has 2 kids and the elder one is a girl. What is the probability that the younger child is also a girl?
2. Alice has 2 kids and one of them is a girl. What is the probability that the other child is also a girl?

• Oshan says:

for Q 21.

The cases considered are for exactly 1, 2 and 3 sixes respectively. however, the question states “atleast” 1,2 and 3 sixes. I am not sure if this would make a difference in the final answer, but this seems like a mistake. Kindly clarify in case I am missing something here..

• Andrew Morris says:

The answer to question 7 must be wrong, because it is greater than 1.
The English grammar through these questions also has many mistakes, e.g. “are common” should be “are in common”.

• [email protected] says:

Hi!

Could you please clarify the following:

Q7 – The answer A) can not be the correct one since it is > 1 (see Q6 🙂 ). So, D.) should be the right choice
Q28 – it is the Binomial schema, isn’t it? Probability of success (p) = 0.7, q = 0.3, n = 3 and m = 2. Then, answer is 3 * 0.7^2 * 0.3
= 0.44. The choice C.) is incorrect since 0.7*0.3*0.7 and 0.3*0.7*0.7 have to be considered as well.

Thanks.

• [email protected] says:

Also, regarding the Q2 – it is well-known “Boy or Girl paradox” (https://en.wikipedia.org/wiki/Boy_or_Girl_paradox). And both answers – C.) and A.) – are correct depending on the randomizing procedure.

• Aditya Royal Matturi says:

The answer for question 7 is (1/52 c 4)*(48 c 4/52 c 4) please correct that

• Dishashree Gupta says:

Yes, the answer for Q7 is not mentioned. We have removed it from the final calculation.

• Soham Lawar says:

Hello ,
I have following queries ;
1) I request you to elaborate more for question 9
2) In question 19 range of test scores is 18 to 24 and in the explanation you have explained for range 20 to 26
3) In the explanation of question number 40 I think ‘since the probabilities are continuous, the probabilities form a distribution function ‘ is correct
Regards,
Soham

• Dishashree Gupta says:

1) Elaborating on Q9, what I wanted you to calculate is the probability that when you throw a dice 6 times, you should get 1,2,3,4,5,6 in some order.
2) Q19 – this has been rectified.
3) yes, so in case of a distribution function, the probability of a random variable being exactly equal to a particular value is 0. We can only calculate the probability of a random variable being in a range.

• potterhead says:

The answer for question 7 should be D – None of these.
None of the options is a probability value.

• Dishashree Gupta says:

Yes, it has been removed from the final score calculation

• Santhosh says:

Q28 & Q33 both looks similar to me as both are having binomial outputs .
Could you please explain why is it different for Q27 and why cant we use Binomial distribution for this unlike Q33

• Aditya Royal Matturi says:

question 23 is invalid because there can be an infinite number of ways of throwing a coin, though to satisfy the condition there is only one possible way, there are infinite total outcomes for that test case so I guess1/infinite=~0

• Aditya Royal Matturi says:

• [email protected] says:

Hi!

Could you please clarify the following:

Q7 – The answer A) can not be the correct one since it is > 1 (see Q6 ? ). So, D.) should be the right choice
Q28 – it is the Binomial schema, isn’t it? Probability of success (p) = 0.7, q = 0.3, n = 3 and m = 2. Then, answer is 3 * 0.7^2 * 0.3
= 0.44. The choice C.) is incorrect since 0.7*0.3*0.7 and 0.3*0.7*0.7 have to be considered as well.

Thanks.

• hanna yang says:

i think the explanation of question 4 “P(AꓵCc) will be only P(A).”is wrong. it shoud be P(AꓵCc) will be P(A-C), If AꓵC =Φ,then the explanation”P(AꓵCc) will be only P(A).”must be right..

• Stu says:

I believe the correct answer to 7 is B not A. Your reasoning is correct, but I think you made an error in reducing the fraction.

• Dishashree Gupta says:

We removed 7 from the calculation due to discrepancy !

• M Zakaria says:

Regarding Q4. I think we should state explicitly that A, B, and C are independent. This way P(AUBUC) = P(A) + P(B) + P(C). otherwise we need to subtract P(A and B), P(A and C), and P(B and C) from the result.

• Mustafa V says:

Q7 answer seems incorrect. The probability value greater than 1.

• Dishashree Gupta says:

Yes there was an issue with it. It had been removed from the final score calculation

• Mustafa V says:

Q2. Answer should be 0.5. Can be reasoned in 2 ways. 1st child & 2nd child are independent events and it is mentioned that population is balanced, p=.5. Also since 1st child is girl so of the possibilities only gb and gg valid, again p = 0.5.

• Mustafa V says:

Hi,

Regarding Q10 – the solution I have is as follows:-

Jack & Jill in one section – number of ways is (58C18)*(40C20) [A]
Total number of ways is (60C20)*(40C20) [B]
Thus the required probability is A/B which is 19/177.

But this option is not there at all.

Can you review my solution?

• Dishashree Gupta says:

Can you give more detail on your A and B calculation ?

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• Nimrod Ifrach says:

Hi,
Im not sure the answer to q28 is correct. 0.147 is the probability of only 1 case that can happen, but there are 3 possible cases:

So actually its 0.147*3=0.44

• Dishashree Gupta says:

Both Q7 and 28 were removed from the final score calculation due to discrepancy !

• Sarah Nogueira says:

For Q11, the solution says:
“If coin A is selected then the number of times the coin would be tossed for a guaranteed Heads is 2”.
This does not seem right since the probability of getting heads with coin A is 1/2, and each toss is independent, we could toss coin A a hundred times and never observe heads…

• Dishashree Gupta says:

Hi Sarah,

If the probability of getting a heads from a coin is 1/2, can you really say that you would get a heads if you throw the coin twice ?

• MJ says:

No we can’t, thats why I think the solution is flawed.