If a six sided die is rolled three times, what is the probability of

The probability of getting at least 1 even and 2 odd number: 3/6*3/6*3/6= 1/2*1/2*1/2=1/8
Also, number of ways of rolling a six sided die like: (EvenOddOdd, Odd,Even,Odd, OddOddEven)—3 ways —> 3*1/8

The probability of getting at least 1 odd and 2 even numbers:
3/6*3/6*3/6= 1/8
Also, number of ways of rolling a six sided die like: (EvenOddEven, Even,Even,Odd, OddEvenEven)—3 ways—> 3*1/8

3/8+3/8= 3/4
The answer choice is D

Posted from my mobile device

Dear lacktutor

What you appear to have failed to take into account is - the number of ways you can choose the one die that would be either even or odd, i.e. 3C1

Taking that into account, your answer should be as follows:

3C1*3/6*3/6*3/6 (for 1 odd and 2 even) and 3C2*3/6*3/6*3/6 (for 2 odd and 1 even)

3*1/8 + 3*1/8 = 6/8 or 3/4, and hence, (D) is the correct answer choice

PLEASE HIT KUDOS IF YOU LIKE MY SOLUTION

Total outcomes = 6*6*6= 216
If all the numbers are odd 3*3*3 = 27
If all the numbers are even 3*3*3= 27
So sum of all even and all odd outcomes = 54
Probability of all odd or all even 54/216 = 1/4
Hence probability of at least one even and one add = 3/4
Hence D

Posted from my mobile device

Yes, you’re right. I didn’t add the number of combinations within Even and Odd.
Thanks

we are considering cases w/o replacement
so atleast even ; 3/6*3/6*3/6 ; 1/8
and odd ; 1/8
total cases ; 1/8+1/8 ; 1/4
1-1/4 ; 3/4
IMO D

Why we are adding the probabilities at last to get the final resultant 3/4 when the question itself ask to calculate the probability of getting atleast one even number and atleast one odd number ?

*AND* is clearly mentioned : so we should perform multiplication, isn’t it ???

Please correct me

Why can’t we approach like this ;

P(Atleast 1 odd ) = 1-P (All Even ) =1-(1/2)^3= 1-(1/8) = 7/8

P(Atleast 1 Even ) =1-P(All Odd ) = 1-(1/2)^3= 1-(1/8) = 7/8

So final probability = 7/8.7/8 = 49/64

Please correct me Bunuel VeritasKarishma

Posted from my mobile device

The probability of two independent events occurring together is given by the product of their probabilities. These two events (at least one odd and at least one even) are not independent. So this method is not correct.
From the total, if you remove the probability of all odd and all even (mutually exclusive events so you just subtract their probabilities one after the other) event, you get that the remaining cases will have at least one odd and one even outcome.

Now it’s pretty clear VeritasKarishma ?

E O E/O . This anagram wheter EOO or EOE cn be permuted in 3!/2! ways = 3 ways.
Further,

getting probability of (E) (O) (E/O) = 1/2 * 1/2 * 6/6 = 1/4 and now we will permute this EOE or EEO in 3 ways so
multiply it with obtained probability = 1/4 * 3 = 3/4 ----> D

Solution:

The phrases “at least one even number” and “at least one odd number” translate as “two even numbers and one odd number” or “one even number and two odd numbers,”. The probability of getting the former is 1/2 x 1/2 x 1/2 x 3C2 = 1/8 x 3 = 3/8 (notice that 3C2 = 3 is the number of ways of arranging two even numbers and one odd number). Similarly, the probability of getting the latter is also 3/8. Therefore, the overall probability is 3/8 + 3/8 = 6/8 = 3/4.

Alternate Solution:

When a six sided die is rolled three times, the total number of possible outcomes is 6 x 6 x 6 = 216.

Notice that if “getting at least one even number and at least one odd number” is not satisfied, then the outcome consists of either all even or all odd numbers. Of the 216 possible outcomes, 3 x 3 x 3 = 27 of them consist of all even numbers and by the same reasoning, another 27 of them consist of all odd numbers. Thus, 27 + 27 = 54 of the outcomes do not meet the requirement of containing at least one even number and at least one odd number. It follows that 216 - 54 = 162 of the outcomes meet this requirement and hence, the probability is 162/216 = 3/4.

Answer: D

Given that a six sided die is rolled three times and We need to find what is the probability of getting at least one even number and at least one odd number?

As we are rolling a dice three times => Number of cases = \(6^3\) = 216

Now, we need to get at least one even AND at least one odd
=> one has to be even, one has to be odd and the third one can be even or odd

In any attempt, Ways to pick Even number = Ways to pick Odd number = 3
(As there are three even and three odd numbers out of the six possible numbers)

There are three rolls so we can pick the roll in which we get an even number in 3C1 ways (As it can come in any of the three rolls) = 3 ways
Similarly, for getting the odd number we can pick that in 2C1 ways (As it can come in any of the remaining 2 rolls) = 2 ways

=> Probability of getting at least one even number and at least one odd number = (Ways to pick even roll * Ways to pick odd roll * Ways to pick Even number * Ways to pick Even number * Ways to pick any number) / 216 = \(\frac{3 * 2 * 3 * 3 * 3 }{ 216}\) = \(\frac{3}{4}\)

So, Answer will be D
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

­why are we adding 1/8 +1/8 shoudnt we multiply since the question is asking AND ?

KarishmaB and Bunuel

­
We add the probabilities of getting all odd and all even because these are separate and mutually exclusive outcomes—they cannot happen at the same time. When calculating the probability for either one event or another to occur, we add their probabilities. Multiplication would be used if we were looking for the probability of both conditions occurring simultaneously in a single outcome, which is not relevant here as the scenarios (all odd and all even) cannot coexist.­