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My approach:

How many times must a fair six sided die be rolled so that there is at least 1/4 probability that a 4 is rolled?

Probability of at least 1/4 that the number 4 is rolled is same as the probability of the at most 3/4 that the number other than 4 is rolled.

First roll -- probability of not getting a number 4 -- 5/6 ~ 16.67 * 5 > 80%. Hence the probability of getting a 4 is less than 20%.

Second roll -- probability of not getting a number 4 -- 5/6 * 5/6 -- 25/36 ~< 75 %. Hence the probability of getting a 4 is approx. greater than 25%.

Hence answer is 2 (B).
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Cornelius
Let n be the number of times the dice is rolled to ensure that the probability of a 4 is at least 1/4.

Thus,
n*(probability of rolling a 4 in one turn)>= 1/4
n*(1/6)>= 1/4
n>6/4---->n>= 1.5
Hence "minimum number of times to ensure that the probability that a 4 is rolled is at least 1/4" >= 1.5 = 2 (since we cannot have 1.5 rolls)

Ans B


By this logic, in 6 throws, you have a probability of getting 4 as 1. I doubt if your approach is correct.
And in 7 and above throws the probability to get 4 will be more than 1. I dont know if I am missing something
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How I look at it is that since the number of throws is lesser, there is lesser chance of all numbers coming up. Hence, 4 has a chance of coming up more than 1/4 times in two throws.

Another way to look at it is that in two throws only two numbers can come up and the chance of 4 coming up once in two trials will be higher than 1/6

4 not 4
not 4 4

1/6*5/6+5/6*1/6 = 25/36>1/4

sagarsingh
Cornelius
Let n be the number of times the dice is rolled to ensure that the probability of a 4 is at least 1/4.

Thus,
n*(probability of rolling a 4 in one turn)>= 1/4
n*(1/6)>= 1/4
n>6/4---->n>= 1.5
Hence "minimum number of times to ensure that the probability that a 4 is rolled is at least 1/4" >= 1.5 = 2 (since we cannot have 1.5 rolls)

Ans B


By this logic, in 6 throws, you have a probability of getting 4 as 1. I doubt if your approach is correct.
And in 7 and above throws the probability to get 4 will be more than 1. I dont know if I am missing something
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Bunuel,

Could you please help us with the solution for this problem? Thanks
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My approach:

For probability of 1( NO. 4 will come) Min no. of throws =6
For probability of 0.25 ( no. 4 will come), Min No. of throws= 6*0.25=1.5
so minimum no. of throws required are 2.
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