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Standard IX

Mathematics

Sum of Probabilities of Favorable and Unfavorable Event

Two dice are ...

Question

Two dice are thrown. The probability that the sum of the points on two dice will be $7,$ is :

\frac{5}{36}

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\frac{6}{36}

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\frac{7}{36}

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\frac{8}{36}

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Solution

The correct option is B $\frac{6}{36}$
The total number of possible ways is $6 \times 6 = 36$
$∴ n (S) = 36$
Let $E$ be the event that sum of the points on two dice is $7$
$∴ E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$
Then $n (E) = 6$
$∴ P (E) = \frac{n (E)}{n (S)} = \frac{6}{36}$

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Similar questions

Q. Two dice are thrown simultaneously. The probability of obtaining total score of seven is
(a) 5/36
(b) 6/36
(c) 7/36
(d) 8/36

Q. Two dice are thrown simultaneously. The probability of obtaining a total score of

8

k / 36

.Find

k

Two dice, one blue and one grey, are thrown at the same time.

(i) Write down all the possible outcomes and complete the following table:

Event:

Sum of two dice

Probability

(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability. Do you agree with this argument?

Q. (i) Complete the following table:

Event: 'Sum on $2$ dice'	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
Probability	$\frac{1}{36}$						$\frac{5}{36}$				$\frac{1}{36}$

(ii) A student argues that there are

11

possible outcomes

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

and

12

. Therefore, each of them has a probability

\frac{1}{11}

. Do you agree with this argument?
Justify your answer.

Q. Two dice are thrown at the same time. Complete the following table.

Event: Sum on 2 dice	Probability
2	1/36
3
4
5
6
7
8	5/36
9
10
11
12	1/36

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Two dice are thrown. The probability that the sum of the points on two dice will be 7, is :

The correct option is B 636The total number of possible ways is 6×6=36∴n(S)=36Let E be the event that sum of the points on two dice is 7∴E={(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}Then n(E)=6 ∴P(E)=n(E)n(S)=636

Two dice are thrown. The probability that the sum of the points on two dice will be $7,$ is :

The correct option is B $\frac{6}{36}$
The total number of possible ways is $6 \times 6 = 36$
$∴ n (S) = 36$
Let $E$ be the event that sum of the points on two dice is $7$
$∴ E = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}$
Then $n (E) = 6$
$∴ P (E) = \frac{n (E)}{n (S)} = \frac{6}{36}$