**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) = ^{11}C_{3}

= (11⋅10⋅9)/(3⋅2⋅1)

= 165

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

n(A) = ^{7}C_{3}

= (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) = ^{7}C_{1 }⋅ ^{4}C_{2}

= 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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