StudentAidUnit 13: Probability — Exercise 13 1
9th Class Mathematics · Unit 13: Probability
13.1.1.Arshad rolls a dice, with sides labelled L, M, N, O, P, U. What is the probability that the dice lands on consonant?
Given
Sample space
S = {L, M, N, O, P, U} ; n(S) = 6
Event (consonants)
A = {L, M, N, P} ; n(A) = 4
Formula
Probability
P(A) = n(A)n(S)
Working
= 46
Result
P(A) = 23
13.1.2.Shazia throws a pair of fair dice. What will be the probability of getting: (i) sum of dots is at least 4. (ii) product of both dots is between 5 to 10. (iii) the difference between both the dots is equal to 4. (iv) number at least 5 on the first dice and the number at least 4 on the second dice.
Given
Sample space
S = {(1,1),(1,2),ldots,(6,6)} ; n(S) = 36
(i) Event: sum \geq 4
A = {(1,3),(1,4),ldots,(6,6)} ; n(A) = 33
Working
(i)
P(A) = n(A)n(S) = 3336
Result
(i)
P(A) = 1112
Given
(ii) Event: product between 5 and 10
B = {(1,5),(1,6),(2,3),(2,4),(2,5),(3,2),(3,3),(4,2),(5,1),(5,2),(6,1)} ; n(B) = 11
Result
(ii)
P(B) = n(B)n(S) = 1136
Given
(iii) Event: difference = 4
C = {(1,5),(2,6),(5,1),(6,2)} ; n(C) = 4
Result
(iii)
P(C) = n(C)n(S) = 436 = 19
Given
(iv) Event: first die \geq 5 and second die \geq 4
D = {(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)} ; n(D) = 4
D has 6 listed elements, so n(D) should be 6, not 4.
Result
(iv)
P(D) = n(D)n(S) = 636 = 16
13.1.3.One alphabet is selected at random from the word "MATHEMATICS". Find the probability of getting: (i) vowel (ii) consonant (iii) an E (iv) an A (v) not M (vi) not T
Given
Sample space
Total letters = 11 ; n(S) = 11
(i) Vowels
A = {A,E,A,I} ; n(A) = 4
Result
(i)
P(A) = 411
Given
(ii) Consonants
B = {M,T,H,M,T,C,S} ; n(B) = 7
Result
(ii)
P(B) = 711
Given
(iii) an E
C = {E} ; n(C) = 1
Result
(iii)
P(C) = 111
Given
(iv) an A
D = {A,A} ; n(D) = 2
Result
(iv)
P(D) = 211
Given
(v) not M
Eset = {A,T,H,E,T,I,C,S} ; n(Eset) = 9
Result
(v)
P(Eset) = 911
Given
(vi) not T
F = {M,A,H,E,M,A,I,C,S} ; n(F) = 9
Result
(vi)
P(F) = 911
13.1.4.Aslam rolled a dice. What is the probability of getting the numbers 3 or 4? Also find the probability of not getting the numbers 3 or 4.
Given
Sample space
S = {1,2,3,4,5,6} ; n(S) = 6
Event
A = {3,4} ; n(A) = 2
Result
P(A) = 26 = 13
Not 3 or 4
P(A') = 1 - P(A) = 1 - 13 = 23
13.1.5.Abdul Hadi labelled cards from 1 to 30 and put them in a box. He selects a card at random. What is the probability that selected card containing: (i) the number 25 (ii) number between 17 to 22 (iii) number at least 20 (iv) number not 27 and 29 (v) number not between 12 – 15
Given
Sample space
n(S) = 30
Result
(i)
A = {25} ; n(A)=1 ⇒ P(A) = 130
(ii)
A = {17,18,19,20,21,22} ; n(A)=6 ⇒ P(A) = 630 = 15
(iii)
A = {20,21,ldots,30} ; n(A)=11 ⇒ P(A) = 1130
(iv)
A = {27,29} ; n(A)=2 ⇒ P(A) = 230 = 115
Not 27 and 29
P(A') = 1 - 115 = 1415
(v)
A = {12,13,14,15} ; n(A)=4 ⇒ P(A) = 430 = 215
Not between 12-15
P(A') = 1 - 215 = 1315
13.1.6.The probability that Ayesha will pass the examination is 0.85. What will be the probability that Ayesha will not pass the examination?
Given
Probability of passing
P(A) = 0.85
Result
Probability of not passing
P(A') = 1 - P(A) = 1 - 0.85 = 0.15
13.1.7.Taabish tossed a fair coin and rolled a fair dice once. Find the probability of the following events: (i) tail on coin and at least 4 on dice. (ii) head on coin and the number 2,3 on dice. (iii) head and tail on coin and the number 6 on dice. (iv) not tail on coin and the number 5 on dice. (v) not head on coin and the number 5 and 2 on dice.
Given
Sample space
S = {(H,1),ldots,(H,6),(T,1),ldots,(T,6)} ; n(S) = 12
Result
(i)
A = {(T,4),(T,5),(T,6)} ; n(A)=3 ⇒ P(A) = 312 = 14
(ii)
A = {(H,2),(H,3)} ; n(A)=2 ⇒ P(A) = 212 = 16
Given
(iii)
A = {(H,5),(T,5)} ; n(A)=2
The event is "head and tail on coin and the number 6 on dice", so A should be {(H,6),(T,6)}, not {(H,5),(T,5)}. The final probability is unaffected (still 2 outcomes out of 12).
Result
(iii)
P(A) = 212 = 16
Given
(iv)
A = {(T,5)} ; n(A)=1
"Not tail" means head, so A should be {(H,5)}, not {(T,5)}. The final probability is unaffected (still 1 outcome out of 12).
Result
(iv)
P(A) = 112
(iv) complement
P(A') = 1 - 112 = 1112
Given
(v)
A = {(H,5),(H,2)} ; n(A)=2
"Not head" means tail, so A should be {(T,5),(T,2)}, not {(H,5),(H,2)}. The final probability is unaffected (still 2 outcomes out of 12).
Result
(v)
P(A) = 212 = 16
(v) complement
P(A') = 1 - 16 = 56
13.1.8.A card is selected at random from a well shuffled pack of 52 playing cards. What will be the probability of selecting: (i) a queen (ii) neither a queen nor a jack
Given
Sample space
n(S) = 52
Result
(i)
A = 4 queens ; n(A)=4 ⇒ P(A) = 452 = 113
Given
(ii)
A = 4 queens and 4 jacks ; n(A)=8 ⇒ P(A) = 852 = 213
P(A') = 1 - 213 = 1113
13.1.9.A card is chosen at random from a pack of 52 playing cards. Find the probability of getting: (i) a jack (ii) no diamond
Given
Sample space
n(S) = 52
Result
(i)
A = 4 jacks ; n(A)=4 ⇒ P(A) = 452 = 113
Given
(ii)
A = 13 diamonds ; n(A)=13 ⇒ P(A) = 1352 = 14
P(A') = 1 - 14 = 34