10th Class Mathematics Unit 12 Probability Numericals with Solutions

12.1.1.Find the sample space for tossing three coins using tree diagram.

Given

Let H = Head, T = Tail

Sample Space S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

Result

n(S) = 8

12.1.2.A teacher randomly selects one boy and one girl from a group of 2 boys (Hamid, Ahmad) and 2 girls (Zainab, Samia). Draw a tree diagram and list the sample space for all possible outcomes.

Sample Space S = {(Hamid, Zainab), (Hamid, Samia), (Ahmad, Zainab), (Ahmad, Samia)}

Result

n(S) = 4

12.1.3.If A is an event of a random experiment such that P(A):P(Ā) = 17:15 and n(S) = 640, then find (i) P(Ā) (ii) n(A)

Given

P(A):P(bar A) = 17:15,quad n(S) = 640

Working

Let P(A) = 17x, P(bar A) = 15x

Formula

P(A) + P(bar A) = 1

17x + 15x = 1 ⇒ 32x = 1 ⇒ x = 132

Result

(i) P(bar A) = 15x = 15 × 132 = 1532

(ii) P(A) = 17x = 17 × 132 = 1732

Working

n(A) = P(A) × n(S) = 1732 × 640 = 17 × 20

Result

n(A) = 340

12.1.4.A coin is tossed thrice. What is the probability of getting two consecutive tails?

Given

Let H = Head, T = Tail

Total outcomes = 2³ = 8; S={HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}

Outcomes with two consecutive tails {HTT, TTH, TTT}

Formula

Probability frac{Favourable{Total = 38

Result

P(two consecutive tails) = 38

12.1.5.A dice is rolled and coin is tossed together. (i) Find sample space by drawing possibility diagram. (ii) Find sample space by sketching tree diagram. (iii) What is a probability of getting a tail and an even number?

(i) Possibility diagram S = {(1,H),(2,H),(3,H),(4,H),(5,H),(6,H),(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}

Result

n(S) = 12

(ii) Tree diagram S = {(1,H),(1,T),(2,H),(2,T),(3,H),(3,T),(4,H),(4,T),(5,H),(5,T),(6,H),(6,T)}

Given

(iii) Even numbers on dice = {2,4,6}

With tail (T) = {(2,T),(4,T),(6,T)}

Favourable outcomes = 3, n(S)=12

Formula

Probability 312 = 14

Result

P(T and even number) = 14

12.1.6.Two unbiased dice are rolled once. (i) Find sample space by sketching tree diagram. (ii) Find sample space by drawing possibility diagram. (iii) Find the probability of getting (a) same number on both dice (b) the product as a prime number (c) the sum as an even number (d) the sum as 13.

(i) Tree diagram n(S) = 36

(ii) Possibility diagram n(S) = 36

Given

(iii)(a) Same number on both dice = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}

Favourable = 6

Result

P = 636 = 16

Given

(iii)(b) Product is prime for: 2, 3, 5

Favourable outcomes = {(1,2),(2,1),(1,3),(3,1),(1,5),(5,1)}, Favourable = 6

Result

P = 636 = 16

(iii)(c) Sum is even (both odd or both even) = 18 outcomes

P = 1836 = 12

(iii)(d) Sum is 13 to not possible (maximum sum = 12)

P = 036 = 0

12.1.7.Three fair coins are tossed together. Find the probability of getting (i) all tails (ii) at least one head (iii) at most two tails (iv) 2 heads (v) at most 2 heads (vi) no head

Given

S = {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}, n(S)=8

(i) All tails to {TTT}

Result

P(all tails)=18

(ii) At least one head to all except TTT, Favourable=7

P(at least one head)=78

(iii) At most two tails to all except TTT, Favourable=7

P(at most two tails)=78

(iv) 2 heads to {HHT,HTH,THH}, Favourable=3

P(2 heads)=38

(v) At most 2 heads to all except HHH, Favourable=7

P(at most 2 heads)=78

(vi) No head to {TTT}, Favourable=1

P(no head)=18

12.1.8.A bag contains 4 red balls, 5 white balls, 6 green balls and 3 black balls. Ali draws a ball at random from the bag. Find the probability that the ball drawn is (i) white (ii) red (iii) not white (iv) not black

Given

Total number of balls = 4+5+6+3 = 18

Result

(i) P(white) = 518

(ii) P(red) = 418 = 29

(iii) Not white to 18-5=13

P(not white) = 1318

(iv) Not black to 18-3=15

P(not black) = 1518 = 56

12.1.9.A number is selected at random from the set of whole numbers 1 to 15, both inclusive. Find the probability that the number selected is: (i) odd (ii) a multiple of 5 (iii) the square of 2 (iv) prime (v) 20

Given

S={1,2,ldots,15}, n(S)=15

(i) Odd numbers to {1,3,5,7,9,11,13,15}

Result

P(odd)=815

(ii) Multiples of 5 to {5,10,15}

P(multiple of 5)=315=15

(iii) Square of 2 to 4

P(square of 2)=115

(iv) Prime numbers to {2,3,5,7,11,13}

P(prime)=615=25

(v) 20 is not in the set

P(20)=015=0

12.1.10.If the probability of an event A is 7/10, then find the probability of the event 'not A'.

Given

P(A)=710

Formula

P(A)+P(bar A)=1

P(not A) = 1-710 = 1010-710 = 310

Result

P(not A) = 310

12.1.11.A dice is rolled twice. Find the probability of having a number greater than 4 on each roll.

Numbers greater than 4 = {5,6}, Favourable=2, Total=6

Result

P(number greater than 4 on one roll) = 26=13

Formula

Since two rolls are independent P(greater than 4 on both rolls) = 13 × 13

Result

P(greater than 4 on each roll) = 19

Unit 12: Probability — Exercise 12 1