10th Class Mathematics Unit 12 Probability Numericals with Solutions

12.3.1.Two numbers are randomly chosen from 1 to 10 with replacement. Find the probability that: (i) both are prime (ii) their product is even

Given

Total numbers = 10 (1 to 10); with replacement, Total outcomes=10 × 10=100

(i) Prime numbers from 1 to 10: 2,3,5,7 (4 numbers); Favourable=4 × 4=16

Result

P(both are prime)=16100=425

Given

(ii) Even numbers={2,4,6,8,10} (5); Odd numbers={1,3,5,7,9} (5)

Product is even except when both are odd; Outcomes both odd=5 × 5=25

Favourable (product is even) = 100-25=75

Result

P(product is even)=75100=frac34

12.3.2.One letter is chosen from the word PUNJAB and another from LAHORE. What is the probability that: (i) both are vowels (ii) one is consonant and other is vowel

Given

PUNJAB Letters={P,U,N,J,A,B} (6); Vowels={U,A} (2); Consonants={P,N,J,B} (4)

LAHORE Letters={L,A,H,O,R,E} (6); Vowels={A,O,E} (3); Consonants={L,H,R} (3)

Total outcomes=6 × 6=36

(i) Favourable=2 × 3=6

Result

P(both are vowels)=636=frac16

(ii) Favourable=(4 × 3)+(2 × 3)=12+6=18

P(one consonant and other vowel)=1836=frac12

12.3.3.A single dice is rolled twice. Find the probability that one roll is a multiple of 3 and the other is 5.

Given

n(S)=6 × 6=36

Multiples of 3 = {3,6}

Favourable outcomes (3,5),(6,5),(5,3),(5,6); n(F)=4

Result

P=n(F)n(S)=436=frac19

12.3.4.Two dice are rolled. Find the probability of getting an odd number on one and a multiple of 2 on other.

Given

n(S)=36; Odd numbers={1,3,5} (3); Multiples of 2={2,4,6} (3)

(odd,even) pairs = 9; (even,odd) pairs = 9

Total favourable outcomes = 9+9=18

Result

P(odd on one and multiple of 2 on other) = 1836=frac12

12.3.5.From a pack of well shuffled cards, two cards are drawn at random one by one with replacement. Find the probability that the first is heart and second is king.

Given

Total cards=52; Hearts=13; Kings=4; with replacement, events are independent

First card is heart P(heart on first draw) = 1352=frac14

Second card is king P(king on second draw) = 452=frac1{13}

Formula

P(heart first and king second) = P(heart first) × P(king second)

Result

=frac14 × frac1{13}=frac1{52}

12.3.6.If two cards are selected from a standard deck of 52 cards without replacement, find the probability that (i) Both are black. (ii) Both are queens. (iii) Both are spades. (iv) Both are diamonds.

Given

Total cards=52; Total ways to draw 2 cards = {}⁵²C₂=52 × 512=1326

(i) Both are black Black cards=26; Favourable ways = {}²⁶C₂=26 × 252=325

Result

P(both black) = 3251326

(ii) Both are queens Queens=4; Favourable ways = {}⁴C₂=4 × 32=6

P(both queens) = 61326=1221

(iii) Both are spades Spades=13; Favourable ways = {}¹³C₂=13 × 122=78

P(both spades) = 781326=117

(iv) Both are diamonds Diamonds=13; Favourable ways = {}¹³C₂=13 × 122=78

P(both diamonds) = 781326=117

12.3.7.Saleem draws two cards one by one without replacement from a well shuffled pack of 52 playing cards. What is the probability that first card is jack and the second card is queen.

Given

Total cards=52; Jacks=4; Queens=4

First card is jack P(jack on first draw) = 452=frac1{13}

Since no replacement cards left=51, Queens still=4

Second card is queen P(queen on second drawmid first is jack) = 451

Formula

P(jack first and queen second) = P(jack first) × P(queen second)

Result

=frac1{13} × frac4{51}=frac4{663}

Unit 12: Probability — Exercise 12 3