# SOLVED PROBABILITY PROBLEMS FOR GRADE 11

Solved Probability Problems for Grade 11 :

Here we are going to see some example problems on probability.

Let S be the sample space associated with a random experiment and A be an event. Let n(S) and n(A) be the number of elements of S and A respectively. Then the probability of the event A is defined as

P(A)  =  n(A)/n(S)

=  Number of cases favourable to A/Exhaustive number of cases in S

Question 1 :

An integer is chosen at random from the first 100 positive integers. What is the probability that the integer chosen is a prime or multiple of 8?

Solution :

Sample space  =  {1, 2, 3, 4, ...........100}

n (S)  =  100

Let "A" be the event of choosing a number is a prime

A  =  {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

n(A)  =  25

P(A)  =  n(A)/n(S)  =  25/100

Let "B" be the event of choosing be a multiple of 8.

B  =  {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96}

n(A)  =  12

P(B)  =  n(B)/n(S)  =  12/100

P(A or B)  =  P(A) + P(B)

=  25/100 + 12/100

=  37 /100

Question 2 :

A bag contains 7 red and 4 black balls, 3 balls are drawn at random. Find the probability that (i) all are red (ii) one red and 2 black.

Solution :

Total number of balls  =  7 red + 4 black

=  11 balls

n(S)  =  11C3

=  (11⋅10⋅9)/(3⋅2⋅1)

=  165

Let "A" be the event of getting 3 balls are in red color.

n(A)  =  7C3

=  (7⋅6⋅5)/(3⋅2⋅1)

=  35

P(A)  =  n(A)/n(S)

=  35/165

P(A)  =  7/33

(ii) one red and 2 black.

Let "B" be the event of getting 1 red and 2 black balls.

n(B)  =  7C⋅ 4C2

= 7 ⋅ 6

=  42

P(B)  =  n(B)/n(S)

=  42/165

P(B)  =  14/55

Question 3 :

A single card is drawn from a pack of 52 cards. What is the probability that (i) the card is an ace or a king (ii) the card will be 6 or smaller (iii) the card is either a queen or 9?

Solution :

Number of cards  =  52

n(S)  =  52

(i) the card is an ace or a king

Let "A" and "B" be the events of getting an ace and king card respectively.

n(A)  =  4 ==> P(A)  =  n(A) / n(S)  =  4/52

n(B)  =  4 ==> P(B)  =  n(B) / n(S)  =  4/52

P(A or B)  =  P(A) + P(B)

=  (4/52) + (4/52)

=  8/52

P(A or B)  =  2/13

(ii) the card will be 6 or smaller

Let "C" and "D" be the events of getting 6 and smaller than 6 respectively.

Out of 13 spade cards, there will be 1 card numbered 6.

Out of 13 diamond cards, there will be 1 card numbered 6.

Out of 13 clever cards, there will be 1 card numbered 6.

Out of 13 heart cards, there will be 1 card numbered 6.

So, there are 4 cards numbered 6.

n(C)  =  4 ==> P(C)  =  n(C) / n(S)  =  4/52

There will be 5 cards in each set of cards.

n(D)  =  16 ==> P(D)  =  n(D) / n(S)  =  16/52

P(C or D)  =  P(C) + P(D)

=  (4/52) + (16/52)

=  20/52

P(C or D)  =  5/13

(iii) the card is either a queen or 9?

Let "E" and "F" be the events of getting queen and a card numbered 9 respectively.

There will be 4 queen cards in 52 cards.

n(E)  =  4  ==>  p(E)  =  n(E)/n(S)  =  4/52

n(F)  =  4 ==>  p(F)  =  n(F)/n(S)  =  4/52

P(C or D)  =  P(C) + P(D)

=  (4/52) + (4/52)

=  8/52

P(E or F)  =  2/13

After having gone through the stuff given above, we hope that the students would have understood, "Solved Probability Problems for Grade 11"

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