Question

Probability
Class 10
Please answer the given 2 questions...​

Answer 1

ANSWER :–

(1) n(S) = 2³ = 8

• Sample space = { HHH , HHT , HTH , THH , HTT , THT ,TTH , TTT }

▪︎ Probability = n(E)/n(S)

(i) P(one head) = 3/8

(ii) P(two head) = 3/8

(iii) P(all heads) = 1/8

(iv) P(at least two head) = 4/8 = ½

(2) Sample space = {2 , 3 , ........ , 101}

▪︎ n(S) = 100

(i) E = a two digit number

▪︎ E = { 10 , 11 ,.........99}

=> n(E) = 90

▪︎ Probability = n(E)/n(S)

=> Probability = 90/100

=> Probability = 9/10

(ii) E = a perfect square number

=> E = {4 , 9 , 16 , 25 , 36 , 49 , 64 , 81 , 100 }

=> n(E) = 9

=> Probability = 9/100

(iii) E = a number divisible by 5

=> E = {5 , 10 , 15 , ..........,100}

=> n{E} = 20

=> Probability = 20/100

=> Probability = ⅕

(iv) E = a prime number less than 100

=> E = {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{E} = 25

=> Probability = 25/100

=> Probability = ¼

Answer 2

Q1.

◘ Given ◘

Three unbiased coins are tossed together.

◘ To Find ◘

The probability of getting  :

(i) one head

(ii) two heads

(iii) all heads

(iv) at least two heads

◘ Solution ◘

When 3 coins are tossed at a time, the possible outcomes are :-

(HHH) or (HHT) or (HTH) or (THH) or (HTT) or (THT) or (TTH) or (TTT), where H is denoted for head and T is denoted for tail.

► Total number of outcomes = 8

▬▬▬▬▬▬▬▬▬▬▬▬

(i) For 1 head, the possible outcomes are :

→ HTT, THT & TTH

Therefore, the probability of getting 1 head = 3/8.

▬▬▬▬▬▬▬▬▬▬▬▬

(ii) For 2 heads, the possible outcomes are :

→ HHT, HTH & THH

Therefore, the probability of getting 2 heads = 3/8.

▬▬▬▬▬▬▬▬▬▬▬▬

(iii) For all heads, the possible outcomes are :

→ HHH

Therefore, the probability of getting all heads = 1/8.

▬▬▬▬▬▬▬▬▬▬▬▬

(iv) For at-least 2 heads, the possible outcomes are :

→ HHH, HHT, HTH & THH

Therefore, the probability of getting at-least 2 heads = 2/8 = 1/4.

____________$\bigstar$_____________

Q2.

◘ Given ◘

A box contains discs which are numbered from 2 to 101.

◘ To Find ◘

If one disc is drawn at random from the box, the probability of getting :

(i) a two-digit number

(ii) a perfect square number

(iii) a number divisible by 5

(iv) a prime number less than 100

◘ Solution ◘

► The total number of outcomes = 101 - 2 + 1 = 100

▬▬▬▬▬▬▬▬▬▬▬▬

(i) The number of two-digit numbers = 10, 11, 12, ..., 98, 99

→ Possible outcomes = 90

Therefore, the probability of getting a two-digit number = 90/100 = 9/10.

▬▬▬▬▬▬▬▬▬▬▬▬

(ii) Perfect squares between 2 and 101 = 4, 9, 16, 25, 36, 49, 64, 81 & 100.

→ Possible outcomes = 9

Therefore, the probability of getting a perfect square number = 9/100.

▬▬▬▬▬▬▬▬▬▬▬▬

(iii) Numbers divisible by 5 are = 5, 10, 15, 20, ...., 90, 95 & 100.

→ Possible outcomes = 20

Therefore, the probability of getting a number divisible by 5 = 20/100 = 1/5.

▬▬▬▬▬▬▬▬▬▬▬▬

(iv) Prime number less than 100 are = 2, 3, 5, 7, 11, ..., 83, 89 & 97.

→ Possible outcomes = 25

Therefore, the probability of getting a a prime number less than 100 = 25/100 = 1/4.

____________$\bigstar$__________ ___