Latest Elitmus Aptitude Question SOLUTION: three dices are thrown what is d probabily to gat atleast one six?

Probality of not getting any six is = (5/6)^3 = 125/216
Therefore Probability of getting at least one six is = 1-125/216 = 91/216

the probabilty of getting no six at all is (5/6)*(5/6)*(5/6)=125/216
now probabilty of getting atleast 1 six is= 1-probabilty of getting no six at all
= 1-125/216
=91/216(Ans)

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

Here There Is Probability Of Getting 1,2,3 Sixs...So (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

Answer: 91/216

P(E)= number of possible events/number of sample space
P(E)=7/216

probablity to get any six when only one dice is thrown=1/6
so,probablity of not getting any six=1-1/6=5/6
now for 3 dice it becomes=(5/6)^3=125/216
therefore probablity of getting atleast one six=1-125/216=91/216.

favorable condition = 7
total condition = 216
probability = 7/216

ans is. 91/216

for no 6 number=5/6*5/6*5/6
125/216
atleast one 6 =1-125/216
=91/216 answer

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Let, u hab to make a 3 digit num using nos. 1 to 6 (repetition allowed),
Total = 6x6x6 =216
Now, by only using nos. 1 to 5, we can make, 5x5x5= 125
No. of 3 digit nums with atleast one 6 are, 216-125= 91,
So, probability will be 91/216.

ans is 36/216=1/6

possible combinations are
(611,612,613,614,615,616) ,(621,622........626),(631,632........636),(641,.........646),(651......656)(661...666)
6+ 6+ 6+ 6+ 6+ 6

=36

alternate
using p&c
6 ( 6 * 6 ) = 36
--- ----

therefore, probability = (36/216) = (1/6)

Correct ans: 91/216

A. There is a total of 6^3=216 combinations if you roll 3 dice. There are 5^2*3=75 combinations that you will get one 6. Thus there is a 75/216=25/72 chance of getting only one 6 when rolling 3 dice.

B. There are 5*3 combinations that you will get 2 6(six). Thus there is a 15/216=5/72 chance of getting a 2 6(six) when rolling 3 dice.

C. There is 1 combination where you will get 3 6s. Thus there is a 1/216 chance you will get 3 6s when rolling 3 dice.

D. There is a 75+15+1/216=91/216 chance of any of them happening.