Latest Elitmus Aptitude Question SOLUTION: what is the probability of getting a six atleast once if three dice are thrown ? 1)5/6*5/6*5/6 2)1/6*5/6*5/6 3)1/6*1/6*1/6 4)none of the above

1-5/6*5/6*5/6=1-125/216=91/216

none of the above

Find the number of cases in which none of the digits show a '6'.

i.e. all three dice show a number other than '6', 5 * 5 *5 = 125 cases.

Total possible outcomes when three dice are thrown = 216.

The number of outcomes in which at least one die shows a '6' = Total possible outcomes when three dice are thrown - Number of outcomes in which none of them show '6'.

= 216 - 125 = 91.

The required probability = 91 / 216
so,answer is none of the above.

Every die has got six sides. Each of the sides is numbered from 1 to 6.

When a single unbiased die is thrown you can have six possible outcomes.

When two dice are thrown simultaneously, the total number of outcomes will be 6 * 6 = 36
Similarly, when three dice are thrown simultaneously, the total number of outcomes will be 6*6*6=216.

We need to find out the number of cases in which at least one of the facing sides shows 6.

At least one means - either one dice or two dice or all three.

Case 1: Let us take the easiest case first - all three dice showing '6' - There is only one such possibility.

Case 2: The number of cases in which two of the dice show 6 and one of them is a different number.
For eg an event like 6 6 5 will be one of the outcomes for case 2.

As two of the dice show '6' , it can happen in only one way. The third die shows a different number, a number other than 6, and it can be any of the 5 other numbers. Therefore, there will 5 possible options i.e. (1, 6, 6), (2, 6, 6), (3, 6, 6), (4, 6, 6), (5, 6, 6).

However, each of these possibilities can have three different arrangements depending upon where the third different digit appears. For example take (1, 6, 6) case - it will have three options (1, 6, 6), (6, 1, 6), (6, 6, 1).

Therefore, the total number of events in which 2 of the dice will show '6' and one will show a different number = 5*3 = 15 cases.

Case 3: When only one of the die shows '6' and the other two show numbers other than '6'.

The die showing '6' has only one option. The other two dice can have any of the '5' options. Therefore, the total number of possibilities = 1*5*5 = 25.

However, the die showing '6' can either be the first die or the second die or the third die.

Therefore, there are a total of 25 * 3 = 75 possibilities.

Total possible outcomes where at least one die shows '6' = Case 1 + Case 2 + Case 3 = 1 + 15 +75 = 91.

Therefore, the required probability = 91/216

3rd option is 3*1/6*5/6*5/6

3rd option is 5/6*5/6*1/6*3

4
1/6+1/6+1/6

for atleast one 6:(1/6*5/6*5/6)+(1/6*1/6*5/6)+(1/6*1/6*1/6)=31/216

Three cases arise:
Case 1: When only one dice shows up a six
This dice can be any of the 1st, 2nd or 3rd dice. Find the probability for these three independent events and add them up to get the total probability
Probability that only 1st dice shows up a six: (probability that first dice shows up a 6) and (probability that second dice shows up other than 6) and (probability that third dice shows up other than 6)
=(1/6)*(5/6)*(5/6)
=25/216
similarly probability that 2nd dice shows up a six: (5/6)*(1/6)*(5/6) = 25/216
And, probability that 3rd dice shows up a six: (5/6)*(5/6)*(1/6) = 25/216
So probability that only one dice shows up a six: (25/216)+(25/216) +(25/216) = 75/216
Case 2:When two dice show up a six
Total number of ways of selecting a pair of dice that show up a six from a set of 3 dice are: 3C2=3
Find the probability of getting six on a pair of dice and multiply it by total number of such possible pairs
Probability of getting a six on a pair of dice = (1/6)*(1/6)*(5/6) = 5/216
So, total probability = 3*(5/216) = 15/216
Case 3: When all dice show up a six
In this case total probability is just (1/6)*(1/6)*(1/6) = 1/216

So total probability of getting at least one six = (75/216) + (15/216) + (1/216) = 91/216