Section 3, Unit 6 Answer wrong

Quant Trading Strategies & Ideas
1

Course Name: Quant Interview Questions Preparation, Section No: 3, Unit No: 6, Unit type: Quiz

I think the answer of the question and even the hints are wrong. 

2

Hi Aditya,



The hint was a little vague and has been changed but the answer is still correct.



You know that the probability of getting a 6 in one dice roll is 1/6. Similarly, the probability of not getting a 6 in one dice roll is 5/6.



For independent events, you multiply the probability of an event's occurrence.

Thus, in the first case, the probability of not getting a 6 when 6 dice are rolled is (5/6)(5/6)(5/6)(5/6)(5/6) i.e. (5/6)^6 = 0.334.

Now, you have calculated the probability of not getting a 6. You know that the total probability of all events is 1. Thus, if you want to find the probability of at least one six, you will use the following formula:- 1 - (Probability of getting a 6 in 6 rolls) = 1 - 0.334 = 0.666.





If you calculate the other scenarios in a similar fashion, you will get the correct answer.





In the second case, you can check two conditions first.

  1. No 6 at all
  2. Exactly one six.

    Then you can deduct the probability values from 1 to get the probability of at least one six in 12 dice rolls.



    For calculating the probability of "No 6 at all", it is the same as previous case, which is (5/6)^12.



    Let us see the probability of exactly one six. 



    Here, consider the case that the first dice roll is 6 and the other 11 are not 6. Thus, for this case, the probability is (1/6) * (5/6)^11.



    Then there will be a case where the second dice roll is 6. This is a different event than the previous case, where it was the first roll.

    But the probability for this case is also (1/6)(5/6)^11.

    In this manner, there are 12 different cases.



    Thus, the total probability will be (12)    (1/6)(5/6)^11



    So the probability that we roll at least 2 sixes in 12 dice rolls is

    (Total probability of all events) - (Probability of no 6) - (Probability of at least one 6)

    = (1 - ((5/6)^12) - ((12)    (1/6)*(5/6)^11)

    = 0.618



    As you can see, this is less than the probability of getting one 6 in 6 rolls.

    You can calculate the third case in a similar fashion.



    Hope this helps.