Page 205 - Maths Skills - 8

Data Handling 203

Solution: Since, two coins are tossed, the sample space would be S = {(HH), (HT), (TH), (TT)}, i.e., 4

(i) Favourable outcomes for (ii) Favourable outcomes for at least
two tails is (TT), i.e., 1 one tail are (HT), (TH), (TT), i.e., 3

1
3
⇒ P(two tails) = . ⇒ P(at least one tail) = .
4 4
(iii) Favourable outcomes for no tail is HH, i.e., 1

1
⇒ P(no tail) = .
4
Example 3: Three coins are tossed. Find the probability of getting:
(i) three heads together (ii) exactly one tail (iii) two heads

Solution: Since three coins are tossed, the total number of possible outcomes will be 2 , i.e., 8. Sample
3
spaces are as follows:
Sample space (s): HHH, HHT, HTH, HTT, THH, THT, TTH, TTT
1
(i) The probability of getting three heads is P(three heads) = Number of favourableoutcomes = .
S 8
3
3
(ii) P(exactly one tail) = . (iii) P(two heads) = .

8 8
 The no. of favourable outcomes  The no. of favourable outcomes
are 3, i.e., (HHT), (HTH), (THH). are 3, i.e., (HHT), (HTH), (THH).

Example 4: Find the probability of getting a number less than 5 in a single throw of a dice.

Solution: A number less than 5 means, 1, 2, 3 or 4.
⇒ Favourable number of outcomes are 4 and S is 6

2
∴ P (a number less than 5) = Favourable outcomes = 4 = .
S 6 3
Example 5: Two die are thrown together. Find the probability of getting.
(i) a sum of 11 (ii) a sum of 9

Solution: We know that the number of possible outcomes are 6 × 6 = 36
(i) a sum of 11 can be possible with (5, 6) and (6, 5), i.e., 2 ways

Hence P (a sum of 11) = 2 = 1 .
36 18
(ii) a sum of 9 can be obtained as (3, 6), (6, 3), (4, 5), (5, 4), i.e., 4 ways.

1
⇒ P (a sum of 9) = 4 = .
36 9

Exercise 12.4

1. Two coins are tossed simultaneously, find the probability of getting
(i) exactly one head (ii) two heads (iii) no heads