byRituparna Nath Content Writer at Study Abroad Exams
Question: Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?
- [1/2]
- [3/4]
- [3/8]
- [5/16]
- [5/6]
‘Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?’ - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide Quantitative Review”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.
Solution and Explanation
Approach Solution 1:
This is a dice sum. As we know a dice has 6 faces. So find the probability we need to first find the odd product. So the odd faces of the dice are 1, 3 and 5.
The probability of odd products = 3∗3=9
Total ways to choose two numbers= 6∗6=36
So to picking number that is even number is 36-9=27
Therefore the probability of picking even product = \(\frac{27}{36}=\frac{3}{4}\)
Thus the probability of getting two numbers whose product is even = \(\frac{3}{4}\)
Hence option B is the correct answer.
Correct Answer: B
Approach Solution 2:
This is a probability sum with complex events. We can use the given below equation to solve this question.
Probability (Complex event) = Probability (Individual Events) * arrangement
We usually represent complex events by a set of alphabets. For instance, if we need to find the probability of getting two heads and a tail when we toss a coin 3 times, then the probability can be represented as P(HHT).
P(HHT) = 1/2 * 1/2 * 1/2 * 3!/2! (3!/2! -----> arrangement of the word HHT)
P(HHT) = 3/8
So to picking number that is even number, we here have two cases:
Case 1. One number is even and one number odd, let us represent this as P(EO)
Case 2. Both numbers are even, let us represent this as P(EE)
For case 1: P(EO) = 3/6 * 3/6 * 2! = ½
For case 2: P(EE) = 3/6 * 3/6 * 2!/2! = ¼
So the equation becomes:
1/2 + 1/4 = ¾
Hence option B is the correct answer.
Correct Answer: B
Approach Solution 3:
The probability in a simultaneous throw of two dice, we have,the number of outcomes be n(S)=(6×6)=36
Let, E be the event that the product of the numbers obtained is even.
So the outcomes are, E ={(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), (3,2),(3,4),(3,6),(4,1),(4,2),(4,3),(4,4)(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3), (6,4),(6,5),(6,6)}
Number of outcomes be 27
that implies P(E) = Total number of outcomes number of favorable outcomes
=27/36
=3/4
Hence, the correct option is B.
Correct Answer: B
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