Sasi Math (Revised for 1024)

Scroll:Probability >> probability >> ps (4488)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

A die is thrown twice. Find the probability of getting a total of 12

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 2

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 11

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 9

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 4

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 5

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 8

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 10

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 3

Answer:_______________

10)

A die is thrown twice. Find the probability of getting a total of 7

Answer:_______________

A die is thrown twice. Find the probability of getting a total of 12

Answer: $\frac{1}{36}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 12.

A = { (6,6) }

n(A) = 1.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{1}{36}$

A die is thrown twice. Find the probability of getting a total of 2

Answer: $\frac{1}{36}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 2.

A = { (1,1) }

n(A) = 1.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{1}{36}$ .

A die is thrown twice. Find the probability of getting a total of 11

Answer: $\frac{1}{18}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 11.

A = { (5,6), (6,5) }

n(A) = 2.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{2}{36}$ = $\frac{1}{18}$

A die is thrown twice. Find the probability of getting a total of 9

Answer: $\frac{1}{9}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 9.

A = { (3,6), (4,5), (5,4), (6,3) }

n(A) = 4.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{4}{36}$ = $\frac{1}{9}$

A die is thrown twice. Find the probability of getting a total of 4

Answer: $\frac{1}{12}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 4.

A = { (1,3), (2,2), (3,1) }

n(A) = 3.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{3}{36}$ = $\frac{1}{12}$ .

A die is thrown twice. Find the probability of getting a total of 5

Answer: $\frac{1}{9}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 5.

A = { (1,4), (2,3), (3,2), (4,1) }

n(A) = 4.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{4}{36}$ = $\frac{1}{9}$

A die is thrown twice. Find the probability of getting a total of 8

Answer: $\frac{5}{36}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 8.

A = { (2,6), (3,5), (4,4), (5,3), (6,2) }

n(A) = 5.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{5}{36}$

A die is thrown twice. Find the probability of getting a total of 10

Answer: $\frac{1}{12}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 10.

A = { (4,6), (5,5), (6,4) }

n(A) = 3.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{3}{36}$ = $\frac{1}{12}$

A die is thrown twice. Find the probability of getting a total of 3

Answer: $\frac{1}{18}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 3.

A = { (1,2), (2,1) }

n(A) = 2.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{2}{36}$ = $\frac{1}{18}$

10)

A die is thrown twice. Find the probability of getting a total of 7

Answer: $\frac{1}{6}$

SOLUTION 1 :

When two dice are thrown, the sample space is

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

n (S) = 6 x 6 = 36.

Let A be the event of getting a sum 7.

A = { (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }

n(A) = 6.

Hence P(A) = $\frac{n(A)}{n(S)}$

= $\frac{6}{36}$ = $\frac{1}{6}$