Sasi Math (Revised for 1024)

Scroll:Probability >> Addition theorem on probability >> ps (4498)

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.

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 4 or 6.

Answer:_______________

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 3 or 6.

Answer:_______________

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 6.

Answer:_______________

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 7.

Answer:_______________

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 8 .

Answer:_______________

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 5.

Answer:_______________

Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 3 nor by 4 .

Answer:_______________

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 9.

Answer:_______________

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 4 or 6.

Answer:_______________

10)

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 3 or 6.

Answer:_______________

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 4 or 6.

Answer: $\frac{8}{25}$

SOLUTION 1 :

Sample space S = { 1, 2, 3, 4, 5, .... 50 }

n(S) = 50.

Let A be the event of getting a number divisible by 4 .

A = { 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 }

n(A) = 12 .

P(A) = n(A) / n(S) = $\frac{12}{50.}$

Let B be the event of getting a number is divisible by 6 .

B = { 6, 12, 18, 24, 30, 36, 42, 48 }

n(B) = 8.

P(B) = n(B) / n(B) = $\frac{8}{50.}$

Since A and B are bot mutually excusive events,

A∩B = { 12, 24, 36, 48 }

n(A∩B) = 4.

P(A∩B) = n(A∩B) / n(S) = $\frac{4}{50.}$

P(A or B) = P(A) + P(B) - P(A∩B)

P(A∪B) = $\frac{12}{50}$ + $\frac{8}{50}$ - $\frac{4}{50}$

= 12 + 8 - 4 / 50

= $\frac{16}{50}$ = $\frac{8}{25}$ .

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 3 or 6.

Answer: $\frac{2}{5}$

SOLUTION 1 :

Sample space S = { 1, 2, 3, 4, 5, .... 50 }

n(S) = 50.

Let A be the event of getting a number divisible by 3 .

A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48 }

n(A) = 16 .

P(A) = n(A) / n(S) = $\frac{16}{50.}$

Let B be the event of getting a number is divisible by 6 .

B = { 6, 12, 18, 24, 30, 36, 42, 48 }

n(B) = 8.

P(B) = n(B) / n(B) = $\frac{8}{50.}$

Since A and B are bot mutually excusive events,

A∩B = { 12, 24, 36, 48 }

n(A∩B) = 4.

P(A∩B) = n(A∩B) / n(S) = $\frac{4}{50.}$

P(A or B) = P(A) + P(B) - P(A∩B)

P(A∪B) = $\frac{16}{50}$ + $\frac{8}{50}$ - $\frac{4}{50}$

= 16 + 8 - 4 / 50

= $\frac{20}{50}$ = $\frac{2}{5}$ .

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 6.

Answer: $\frac{7}{12}$

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 even number in the first die.

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

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

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

n(A) = 18.

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

= $\frac{18}{36}$

Let B be the event of getting the total of 6.

B = { (1,5), (2,4), (3,3), (4,2), (5,1) }

n(B) = 5.

P(B) = n(B) / n(S) = $\frac{5}{36.}$

A ∩ B = { (2,4), (4,2) }

n(A∩B) = 2.

P(A∩B) = n(A∩B) / n(S) = $\frac{2}{36}$

A and B sre not mutually exculusive events.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B)

= $\frac{18}{36}$ + $\frac{5}{36}$ - $\frac{2}{36}$ = 23 - 2 / 36 .

= $\frac{21}{36}$ = $\frac{7}{12}$

P(A∪B) = $\frac{7}{12}$ .

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 7.

Answer: $\frac{7}{12}$

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 even number in the first die.

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

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

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

n(A) = 18.

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

= $\frac{18}{36}$

Let B be the event of getting the total of 7.

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

n(B) = 6

P(B) = n(B) / n(S) = $\frac{6}{36.}$

A ∩ B = { (2,5), (4,3), (6,1)}

n(A∩B) = 3.

P(A∩B) = n(A∩B) / n(S) = $\frac{3}{36}$

A and B are not mutually exculusive events.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B)

= $\frac{18}{36}$ + $\frac{6}{36}$ - $\frac{3}{36}$ = 24 - 3 / 36

= $\frac{21}{36}$ = $\frac{7}{12}$ .

P(A∪B) = $\frac{7}{12}$ .

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 8 .

Answer: $\frac{5}{9}$

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 even number in the first die.

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

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

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

n(A) = 18.

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

= $\frac{18}{36}$

Let B be the event of getting the total of 8.

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

n(B) = 5

P(B) = n(B) / n(S) = $\frac{5}{36.}$

A ∩ B = { (2,6), (4,4), (6,2) }

n(A∩B) = 3.

P(A∩B) = n(A∩B) / n(S) = $\frac{3}{36}$

A and B are not mutually exculusive events.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B)

= $\frac{18}{36}$ + $\frac{5}{36}$ - $\frac{3}{36}$ = 23 - 3 / 36

= $\frac{20}{36}$ = $\frac{5}{9}$ .

P(A∪B) = $\frac{5}{9}$ .

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 5.

Answer: $\frac{5}{9}$

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 even number in the first die.

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

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

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

n(A) = 18.

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

= $\frac{18}{36}$

Let B be the event of getting the total of 5.

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

n(B) = 4.

P(B) = n(B) / n(S) = $\frac{4}{36.}$

A ∩ B = { (2,3), (4,1) }

n(A∩B) = 2.

P(A∩B) = n(A∩B) / n(S) = $\frac{2}{36}$

A and B are not mutually exculusive events.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B)

= $\frac{18}{36}$ + $\frac{4}{36}$ - $\frac{2}{36}$ = 22 - 2 / 36 .

= $\frac{20}{36}$ = $\frac{5}{9}$

P(A∪B) = $\frac{5}{9}$ .

Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 3 nor by 4 .

Answer: $\frac{4}{9}$

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 the sum of the numbers on the faces is divisible by 3 .

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

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

n(A) = 12.

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

= $\frac{12}{36}$

Let B be the event of getting the total of 4.

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

n(B) = 9.

P(B) = n(B) / n(S) = $\frac{9}{36.}$

A ∩ B = { (6,6) }

n(A∩B) = 1.

P(A∩B) = n(A∩B) / n(S) = $\frac{1}{36}$

A and B sre not mutually exculusive events.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B)

= $\frac{12}{36}$ + $\frac{9}{36}$ - $\frac{1}{36}$ = 21 - 1 / 36

= $\frac{20}{36}$ = $\frac{5}{9.}$

P(A∪B) = $\frac{5}{9.}$

Probability of getting the sum of the numbers on the faces is either divisible.

by 3 or divisible by 4 = P(A∪B) = $\frac{5}{9}$ .

Probability of getting the sum of the numbers on the faces is neither divisible by 3 nor divisible by 4 = P(A∪B)â€›

= 1 - P(A∪B) ⇒ ( p(Aâ€›) = 1 - P(A) Required probabability is $\frac{4}{9}$ ).

= 1 - $\frac{5}{9}$ = $\frac{4}{9}$ .

If a die is rolled twice, find the probability of getting an even number in the first time or a total of 9.

Answer: $\frac{5}{9}$

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 even number in the first die.

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

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

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

n(A) = 18.

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

= $\frac{18}{36}$

Let B be the event of getting the total of 9.

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

n(B) = 4

P(B) = n(B) / n(S) = $\frac{4}{36.}$

A ∩ B = { (4,5), (6,3) }

n(A∩B) = 2.

P(A∩B) = n(A∩B) / n(S) = $\frac{2}{36}$

A and B are not mutually exculusive events.

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = P(A) + P(B) - P(A∩B)

= $\frac{18}{36}$ + $\frac{4}{36}$ - $\frac{2}{36}$ = 22 - 2 / 36

= $\frac{20}{36}$ = $\frac{5}{9}$ .

P(A∪B) = $\frac{5}{9}$ .

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 4 or 6.

Answer: $\frac{8}{25}$

SOLUTION 1 :

Sample space S = { 1, 2, 3, 4, 5, .... 50 }

n(S) = 50.

Let A be the event of getting a number divisible by 4 .

A = { 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 }

n(A) = 12 .

P(A) = n(A) / n(S) = $\frac{12}{50.}$

Let B be the event of getting a number is divisible by 6 .

B = { 6, 12, 18, 24, 30, 36, 42, 48 }

n(B) = 8.

P(B) = n(B) / n(B) = $\frac{8}{50.}$

Since A and B are bot mutually excusive events,

A∩B = { 12, 24, 36, 48 }

n(A∩B) = 4.

P(A∩B) = n(A∩B) / n(S) = $\frac{4}{50.}$

P(A or B) = P(A) + P(B) - P(A∩B)

P(A∪B) = $\frac{12}{50}$ + $\frac{8}{50}$ - $\frac{4}{50}$

= 12 + 8 - 4 / 50

= $\frac{16}{50}$ = $\frac{8}{25}$ .

10)

One number is chosen randomly from the integers 1 to 50. Find the probabllity that it is divisible by 3 or 6.

Answer: $\frac{2}{5}$

SOLUTION 1 :

Sample space S = { 1, 2, 3, 4, 5, .... 50 }

n(S) = 50.

Let A be the event of getting a number divisible by 3 .

A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48 }

n(A) = 16 .

P(A) = n(A) / n(S) = $\frac{16}{50.}$

Let B be the event of getting a number is divisible by 6 .

B = { 6, 12, 18, 24, 30, 36, 42, 48 }

n(B) = 8.

P(B) = n(B) / n(B) = $\frac{8}{50.}$

Since A and B are bot mutually excusive events,

A∩B = { 12, 24, 36, 48 }

n(A∩B) = 4.

P(A∩B) = n(A∩B) / n(S) = $\frac{4}{50.}$

P(A or B) = P(A) + P(B) - P(A∩B)

P(A∪B) = $\frac{16}{50}$ + $\frac{8}{50}$ - $\frac{4}{50}$

= 16 + 8 - 4 / 50

= $\frac{20}{50}$ = $\frac{2}{5}$ .