Two Dice are Tossed Once. The Probability of Getting an Even Number GMAT Problem Solving

Question: Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is

\(\frac{1}{36}\)
\(\frac{3}{36}\)
\(\frac{11}{36}\)
\(\frac{20}{36}\)
\(\frac{23}{36}\)

Correct Answer: D

Solution and Explanation:
Approach Solution 1:
The problem statement states that:
Given:

Two dice are tossed once.

Find out:

The probability of getting an even number at the first die or a total of 8.

Based on probability, we can solve it as
Considering “OR” probability:
Let us consider that Events A and B are independent.
The probability that Event A OR Event B occurs can be stated as:
P(A or B) = P(A)+P(B) − P(A and B)
This is basically the same as the 2 overlapping sets formula:
{total number of items in groups A or B} = {number of items in group A} + {number of items in group B} - {number of items in A and B}.
It is required to note that if events are mutually exclusive then
P(A and B)=0 and the formula simplifies to:
P(A or B) = P(A)+P(B)
Also, it is required to note that when we say "A or B occurs" we include three possibilities:
A occurs and B does not occur;
B occurs and A does not occur;
Both A and B occur.
Now, we will consider ‘AND” probability:
When two events are independent, the probability of both occurring is the product of the probabilities of the individual events:
P(A and B)=P(A)∗P(B)
This is basically the same as the Principle of Multiplication:
An event can occur in m ways and a second can occur independently of the first in n ways
In this case, the two events can occur in mn ways.

Now let us consider the original questions, keeping the probability in mind:
Two dice are tossed once. The probability of getting an even number at the first die or a total of 8 is
A. \(\frac{1}{36}\)

B. \(\frac{3}{36}\)

C. \(\frac{11}{36}\)

D. \(\frac{20}{36}\)

E. \(\frac{23}{36}\)

The probability of getting an even number at the first die is 1/2 (as the probability of even = the probability of odd = 1/2);
The probability of getting a total of 8 is 5/6^2, as there are 5 different favorable scenarios: (2,6), (6,2), (3,5), (5,3) and (4,4);
The probability of getting an even number at the first die AND a total of 8 is 3/6^2 (from the above case);

Hence, The probability of getting an even number at the first die OR a total of 8 is \(\frac{1}{2}+\frac{5}{36}-\frac{3}{36}=\frac{20}{36}\).

Approach Solution 2:
The problem statement informs that:
Given:

Two dice are tossed once.

Find out:

The probability of getting an even number at the first die or a total of 8.

Total possible outcomes for two dice: 36

Let’s consider case 1
probability of obtaining an even number on the first die
The first die can have 2,4,6
The second die can have any of 1,2,3,4,5,6
The favorable outcomes are 3x6 = 18
Hence, the probability = \(\frac{18}{36}\)

Let us consider case 2
probability of obtaining a sum of 8 (2,6)(6,2)(3,5)(5,3)(4,4)
The probability=\(\frac{5}{36}\)
Now both of the above cases have some cases in common with them
i.e. when the first die has an even number and the sum is also 8
There are 3 cases of this kind (2,6) (6,2) (4,4)
The probability is =\(\frac{3}{36}\)
Also P(A or B)=P(A) + P(B) - P(A & B)
Therefore, we have P(even or sum of 8) = \(\frac{18}{36}+\frac{5}{36}-\frac{3}{36}\) =\(\frac{20}{36}\)
Hence, the probability of getting an even number at the first die or a total of 8 = \(\frac{20}{36}\)

Approach Solution 3:
The problem statement implies that:
Given:

Two dice are tossed once.

Find out:

The probability of getting an even number at the first die or a total of 8.

Since we are rolling two dice, the number of cases = 6² = 36

Case of getting an even number in the first dice
The First dice can get even numbers in 3 ways (2, 4, 6)
The Second dice can get any of the 6 numbers

Possible cases for the two rolls _ _ are 3 * 6 = 18 ways

Case of getting a Sum of 8
We can get the sum of 8 in the following ways
(2, 6), (3, 5), (4, 4), (5, 3), (6, 6)
But all cases except (3, 5) and (5, 3) are regarded above when we took the first dice roll as an even number => 2 ways
=> Total ways = 18 + 2 = 20 ways
Therefore, the probability of getting an even number at the first die or a total of 8 = \(\frac{20}{36}\)

“Two dice are tossed once. The probability of getting an even number”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. This topic has been taken from the book “Kaplan GMAT Math Workbook”. The GMAT Problem Solving questions help the candidates to evaluate facts and solve numerical problems. The candidates can go through GMAT Quant practice papers to practice several sorts of questions that will help them to enhance their mathematical understanding.

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Two Dice are Tossed Once. The Probability of Getting an Even Number GMAT Problem Solving

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