Sasi Math (Revised for 1024)

Scroll:Probability >> probability >> saq (4491)

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.

20 cards are numbered from 1 to 20 . One card is grawn at randon, What is the probablity that the number on the card is a multiples of 4

Answer:_______________

A number is selected at random from integers 1 to 100. Find the probability that it is a perfect square

Answer:_______________

A number is selected at random from integers 1 to 100. Find the probability that it is not a perfect cube.

Answer:_______________

For a sightseeing trip, a tourist selects a country randomly from Argentina, Bangladesh, China, Angola, Russia and Algeria. What is the probability that the name of the selected country will begin with A

Answer:_______________

There dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.

Answer:_______________

A two digit number is formed with the digits 3, 6 and 9. Find the probability that the number so formed is greater than 69 (repetition of digits is not allowed).

Answer:_______________

A two digit number is formed with the digits 2, 4 and 7. Find the probability that the number so formed is greater than 47 (repetition of digits is not allowed).

Answer:_______________

20 cards are numbered from 1 to 20 . One card is grawn at randon, What is the probablity that the number on the card is not a multiples of 6

Answer:_______________

Two dice are rolled and the product of the outcomes (numbers) are found. What is the probability that the product so found is a prime number

Answer:_______________

10)

20 cards are numbered from 1 to 20 . One card is grawn at randon, What is the probablity that the number on the card is a multiples of 4

Answer:_______________

20 cards are numbered from 1 to 20 . One card is grawn at randon, What is the probablity that the number on the card is a multiples of 4

Answer: $\frac{1}{4}$

SOLUTION 1 :

Sample space , n(S) = 20.

Let A be the event of drawing a card, which is multiple of 4.

A = { 4, 8, 12, 16, 20 }

n(A) = 5

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

= $\frac{5}{20}$ = $\frac{1}{4}$

A number is selected at random from integers 1 to 100. Find the probability that it is a perfect square

Answer: $\frac{1}{10}$

SOLUTION 1 :

Sample space, S = { 1,2,3, ...... 100 }

n(S) = 100.

Let A be the event of choosing a perfect square number

A = { 1, 4, 9, 16, 25, 36, 49, 64, 81, 100 }

n(A) = 10.

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

= $\frac{10}{100}$ = $\frac{1}{10}$

A number is selected at random from integers 1 to 100. Find the probability that it is not a perfect cube.

Answer: $\frac{24}{25}$

SOLUTION 1 :

Sample space, S = { 1,2,3, ...... 100 }

n(S) = 100.

Let B be the event of choosing a number, which is not a perect cube.

Perfect cube numbers from 1 to 100 is C = { 1, 8, 27, 64}.

n(C) = 4.

n(B) = n(S) - n(C)

n(B) = 100 - 4 = 96.

n(B) = 96.

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

= $\frac{96}{100}$ = $\frac{24}{25}$

Answer: $\frac{1}{2}$

SOLUTION 1 :

Sample space, S = { Argentina, Bangaladesh, China, Angola, Russia, Algeria }

n(S) = 6.

Let C be the event of selecting the country will begin with A.

C = { Argentina, Angola, Algeria }.

n(C) = 3.

P(C) = $\frac{n(C)}{n(S)}$

= $\frac{3}{6}$ = $\frac{1}{2}$

There dice are thrown simultaneously. Find the probability of getting the same number on all the three dice.

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

SOLUTION 1 :

When three dice are rolled, the sample space

S = { (1,1,1), (1,1,2), (1,1,3), ... (6,6,6).

S contains 6 x 6 x 6 = 216 outcomes.

Let A be the event of getting the sum of face numbers are same.

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

n(A) = 6.

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

= $\frac{6}{216}$ = $\frac{1}{36}$

A two digit number is formed with the digits 3, 6 and 9. Find the probability that the number so formed is greater than 69 (repetition of digits is not allowed).

Answer: $\frac{1}{3}$

SOLUTION 1 :

Sample space, S = { 36, 39, 63, 69, 93, 96 }

n(S) = 6.

Let A be the event of the number so formed is greater than 69.

A = { 93, 96 }.

n(A) = 2.

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

= $\frac{2}{6}$ = $\frac{1}{3}$

A two digit number is formed with the digits 2, 4 and 7. Find the probability that the number so formed is greater than 47 (repetition of digits is not allowed).

Answer: $\frac{1}{3}$

SOLUTION 1 :

Sample space, S = { 24, 27, 42, 47, 72, 74 }

n(S) = 6.

Let A be the event of the number so formed is greater than 47.

A = { 72, 74 }.

n(A) = 2.

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

= $\frac{2}{6}$ = $\frac{1}{3}$

20 cards are numbered from 1 to 20 . One card is grawn at randon, What is the probablity that the number on the card is not a multiples of 6

Answer: $\frac{17}{20}$

SOLUTION 1 :

Sample space , n(S) = 20.

Let A be the event of drawing a card, which is not a multiple of 6.

B = { 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 20 }.

n(B) = 17

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

= $\frac{17}{20}$

Two dice are rolled and the product of the outcomes (numbers) are found. What is the probability that the product so found is a prime number

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 two digit number formed is a prime number.

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

n(A) = 6.

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

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

10)

20 cards are numbered from 1 to 20 . One card is grawn at randon, What is the probablity that the number on the card is a multiples of 4

Answer: $\frac{1}{4}$

SOLUTION 1 :

Sample space , n(S) = 20.

Let A be the event of drawing a card, which is multiple of 4.

A = { 4, 8, 12, 16, 20 }

n(A) = 5

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

= $\frac{5}{20}$ = $\frac{1}{4}$