
Two different dice are tossed together. Find the probability
(i). that the number on each die is even
(ii). that the sum of numbers appearing on the two dice is 5.
A. \[\dfrac{1}{3}\]and \[\dfrac{1}{8}\]
B. \[\dfrac{1}{2}\]and \[\dfrac{1}{7}\]
C. \[\dfrac{1}{5}\]and \[\dfrac{1}{6}\]
D. \[\dfrac{1}{4}\]and \[\dfrac{1}{9}\]
Answer
510.6k+ views
Hint: This is a question of probability. To solve this question first we will obtain total possible outcomes of two dice. From that we will count the number of possibilities where both the dice show an even number. Dividing that number with total outcomes we will get the answer. Similarly we will count the number of times the sum of the number both of the dice appearing is 5 and dividing that number by total outcomes we will get its probability value
Complete step-by-step answer:
Two different dice are tossed together.
Each dice has 6 sides containing 6 numbers i.e. 1, 2, 3, 4, 5 and 6
So the total probability of pairs to be appeared is $ 6 \times 6 = 36 $
Case-1
The favorable outcomes where the number of each dice is even are =
{(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
Hence, the probability of each dice appearing even number = number of favorable outcomes / total outcomes
i.e. the probability of each dice appearing even number $ = \dfrac{9}{{36}} = \dfrac{1}{4} $
Again, the favorable outcomes where the sum of the numbers appearing on two each dice is 5 are =
{(1, 4), (2, 3), (3, 2), (4, 1)}
i.e. the probability that the sum of numbers appearing on both dice is 5 = number of favorable outcomes / total outcomes
Hence, the probability that the sum of numbers appearing on both dice is 5 $ = \dfrac{4}{{36}} = \dfrac{1}{9} $
So, the correct answer is “Option D”.
Note: Probability is concerned with numerical description of how likely an event is to occur and how likely it is that a proposition is true. It is always between 0 and 1.
The formula of probability in simple language is, number of favorable outcomes/total outcomes.
You should practice similar questions of probability related to cards, dice, coins, balls etc.
Be cautious while making the table for total outcomes of two dice.
Complete step-by-step answer:
Two different dice are tossed together.
Each dice has 6 sides containing 6 numbers i.e. 1, 2, 3, 4, 5 and 6
Dice | 1 | 2 | 3 | 4 | 5 | 6 |
1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
So the total probability of pairs to be appeared is $ 6 \times 6 = 36 $
Case-1
The favorable outcomes where the number of each dice is even are =
{(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
Hence, the probability of each dice appearing even number = number of favorable outcomes / total outcomes
i.e. the probability of each dice appearing even number $ = \dfrac{9}{{36}} = \dfrac{1}{4} $
Again, the favorable outcomes where the sum of the numbers appearing on two each dice is 5 are =
{(1, 4), (2, 3), (3, 2), (4, 1)}
i.e. the probability that the sum of numbers appearing on both dice is 5 = number of favorable outcomes / total outcomes
Hence, the probability that the sum of numbers appearing on both dice is 5 $ = \dfrac{4}{{36}} = \dfrac{1}{9} $
So, the correct answer is “Option D”.
Note: Probability is concerned with numerical description of how likely an event is to occur and how likely it is that a proposition is true. It is always between 0 and 1.
The formula of probability in simple language is, number of favorable outcomes/total outcomes.
You should practice similar questions of probability related to cards, dice, coins, balls etc.
Be cautious while making the table for total outcomes of two dice.
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