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If a six-sided die is rolled three times, what is the proba
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13 Jul 2020, 10:04
13 Jul 2020, 10:04
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If a six-sided die is rolled three times, what is the probability that the die will land on an even number exactly twice and on an odd number exactly once?
A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 7/8
Kudos for the right answer and explanation
A. 1/8
B. 1/4
C. 3/8
D. 1/2
E. 7/8
Kudos for the right answer and explanation
Re: If a six-sided die is rolled three times, what is the proba
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13 Jul 2020, 21:07
13 Jul 2020, 21:07
1
Quote:
If a six-sided die is rolled three times, what is the probability that the die will land on an even number exactly twice and on an odd number exactly once?
Step 1: Understanding the question
We require exactly two Even and exactly one Odd, when a dice is rolled thrice.
There are three possible outcomes:
E,E,O Or
E,O,E Or
O,E,E
Step 2: Calculation
We know Probability = \(\frac{Desired number of outcome }{ Total Number of outcome}\)
Getting E,E,O has probability = \(\frac{1}{2} * \frac{1}{2 }* \frac{1}{2} = \frac{1}{8}\)
Similarly getting E,O,E and O,E,E each has probability of \(\frac{1}{8}\)
Required probability = \(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}\)
(C) is correct.
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Re: If a six-sided die is rolled three times, what is the proba
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16 Oct 2022, 03:42
16 Oct 2022, 03:42
Given that a fair 6-sided die is rolled three times and We need to find what is the probability that the die will land on an even number exactly twice and on an odd number exactly once?
As we are rolling three dice => Number of cases = \(6^3\) = 216
Now we have three places in these 3 tosses _ _ _
First of all let's find the toss in which we will get an odd number. This can happen by selecting 1 place out of 3 where odd number will come.
=> 3C1 = 3 ways
Now, Probability of getting an odd number in any toss = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As 3 out of the 6 possible outcomes are odd)
Similarly, P of getting an even number = \(\frac{1}{2}\)
=> Probability that the die will land on an even number exactly twice and on an odd number exactly once = Place of the even number * Probability of getting an odd number * Probability of getting an even number * Probability of getting an even number = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
As we are rolling three dice => Number of cases = \(6^3\) = 216
Now we have three places in these 3 tosses _ _ _
First of all let's find the toss in which we will get an odd number. This can happen by selecting 1 place out of 3 where odd number will come.
=> 3C1 = 3 ways
Now, Probability of getting an odd number in any toss = \(\frac{3}{6}\) = \(\frac{1}{2}\) (As 3 out of the 6 possible outcomes are odd)
Similarly, P of getting an even number = \(\frac{1}{2}\)
=> Probability that the die will land on an even number exactly twice and on an odd number exactly once = Place of the even number * Probability of getting an odd number * Probability of getting an even number * Probability of getting an even number = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Dice Rolling Probability Problems
