StudentAidUnit 12: Probability — Exercise 12 2
10th Class Mathematics · Unit 12: Probability
12.2.1.If A and B are two events such that P(A) = 1/4, P(B) = 1/2 and P(A∩B) = 1/8, find P(A∪B).
Formula
P(Acup B)=P(A)+P(B)-P(Acap B)
Working
= frac14+frac12-frac18 = frac28+frac48-frac18
=frac68-frac18 = frac58
Result
P(Acup B)=frac58
12.2.2.In an apartment, selecting a house from door numbers 1 to 50 randomly, find the probability of getting the door number of the house to be an even number or a perfect square number.
Given
S={1,2,ldots,50}, n(S)=50
A = Even numbers
n(A)=25
B = Perfect square numbers
B={1,4,9,16,25,36,49}, n(B)=7
Acap B = even perfect squares = {4,16,36}, n(Acap B)=3
Formula
P(Acup B) = n(A)n(S)+n(B)n(S)-n(Acap B)n(S)
Working
= 2550+750-350 = 25+7-350
Result
P(Acup B) = 2950
12.2.3.The probability of a team winning any match is 3/10 and the probability of losing any match is 2/10. What is the probability that (i) the team wins or loses a particular match. (ii) the team neither wins nor loses a match.
Given
P(W)=310, P(L)=210
Formula
(i)
P(Wcup L)=P(W)+P(L) (W,L mutually exclusive)
Working
=310+210=510=frac12
Result
P(team wins or loses)=frac12
Formula
(ii)
P(neither wins nor loses) = 1-P(Wcup L)
=1-frac12=frac12
Result
P(team neither wins nor loses)=frac12
12.2.4.In a single throw of two dice, find the probability of having sum of 7 or 11.
Given
n(S)=6 × 6=36
Sum = 7
{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)}, n=6
Sum = 11
{(5,6),(6,5)}, n=2
Favourable = 6+2=8
Result
P(Sum=7 or 11)=836=frac29
12.2.5.Find the probability of getting a sum of 5 or 7 in a throw of two dice.
Given
n(S)=36
Sum = 5
{(1,4),(2,3),(3,2),(4,1)}, n=4
Sum = 7
n=6
Favourable=4+6=10
Result
P(sum=5 or 7)=1036=518
12.2.6.A card is taken out at random from a standard pack of 52 cards. Find the probability of taking out. (i) A king or a Jack. (ii) Neither a king nor a Jack.
Given
Total cards = 52
(i)
Kings=4, Jacks=4, Favourable=4+4=8
Result
P(King or Jack) = 852=213
(ii)
Neither a king nor a Jack to 52-8=44
P(Neither King nor Jack) = 4452=1113
12.2.7.A dice is thrown twice. What is the probability that at least one of the two throws comes up with number 3.
Given
n(S)=6 × 6=36
No '3' in one throw = 5 outcomes
No '3' in both throws = 5 × 5=25
Favourable outcomes (at least one '3') = 36-25=11
Result
P(at least one '3') = 1136
12.2.8.There are 15 cards in a bag marked as 1, 2, 3, ...., 15. Find the probability of picking a card at random, the number written on which is a multiple of 5 or of 7.
Given
S={1,2,ldots,15}, n(S)=15
Multiples of 5
A={5,10,15}, n(A)=3
Multiples of 7
B={7,14}, n(B)=2
Common multiple (5 and 7)
35notin S, so Acap B=emptyset ⇒ n(Acap B)=0
Formula
n(Acup B)=n(A)+n(B)-n(Acap B)=3+2-0=5
Result
P(multiple of 5 or 7) = 515=frac13
12.2.9.Two fair coins are tossed once. What is the probability of getting at least one head or two heads.
Given
S={HH,HT,TH,TT}, n(S)=4
At least one head or two heads = {HH,HT,TH}
Favourable outcomes = 3
Result
P(at least one head or two heads) = frac34
12.2.10.At a busy intersection, 50% of vehicles turn right, 30% turn left and 20% go straight. What is the probability that a randomly selected vehicle turn left or right?
Given
P(R)=0.50, P(L)=0.30, P(S)=0.20
Formula
P(Lcup R)=P(L)+P(R) (L, R mutually exclusive)
Working
=0.30+0.50
Result
P(turn left or right) = 0.80
12.2.11.Two fair coins are tossed. What is the probability of getting either two heads or two tails?
Given
S={HH,HT,TH,TT}, n(S)=4
Favourable outcomes = {HH,TT}, n(F)=2
Result
P(two heads or two tails) = n(F)n(S) = frac24=frac12
12.2.12.If A and B are two mutually exclusive events of a random experiment and P(not A) = 0.45, P(A∪B) = 0.65, then find P(B).
Given
P(bar A)=0.45 ⇒ P(A)=1-0.45=0.55
Formula
P(Acup B)=P(A)+P(B) (A, B mutually exclusive)
0.65=0.55+P(B)
Result
P(B)=0.65-0.55=0.10
12.2.13.If P(A) = 2/3, P(B) = 2/5, P(A∪B) = 1/3, then find P(A∩B).
Formula
P(Acup B)=P(A)+P(B)-P(Acap B)
frac13=frac23+frac25-P(Acap B)
Working
P(Acap B)=frac23+frac25-frac13 = left(1015+frac6{15}-frac5{15}right)
Result
P(Acap B)=1115
12.2.14.A and B are two events such that, P(A) = 0.42, P(B) = 0.48, and P(A∩B) = 0.16. Find: (i) P(Ā) (ii) P(B̄) (iii) P(A∪B)
Formula
(i)
P(bar A)=1-P(A)
=1-0.42=0.58
Result
P(bar A)=0.58
Formula
(ii)
P(bar B)=1-P(B)
=1-0.48=0.52
Result
P(bar B)=0.52
Formula
(iii)
P(Acup B)=P(A)+P(B)-P(Acap B)
Working
=0.42+0.48-0.16=0.90-0.16
Result
P(Acup B)=0.74