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( b) Favourable outcomes (at least one tail) = HT, TH, TT

3
\ P(getting at least one tail) =
4
Now, the probabilities of these events can be represented on the number line as
shown here.
P(getting exactly P(getting at least
two head) = 1 one tall) = 3
4 4
0 1 1
2
Example 14: A coin is tossed 200 times and the heads appear 120 times. Find the probability of
getting tails in this experiment.

Solution: Let E denote the event of getting heads. Note: If the probability of an event E is
P(E) and the probability of the
Then, n(S) = 200 and n(E) = 120
complement of event E (the event
E
\ P(getting a heads), i.e., P(E) = n() = 120 = 3 that does not occur) is PE () , the
S
n() 200 5 sum of both these probabilities is
Here, getting heads and tails are complementary always equal to 1.
events. That is, PE () + () = 1.
PE
3 2
1
E
1
\ P(getting a tails) = P()E = − P() = − =
5 5
Example 15: A glass jar contains 6 red, 5 green, 4 blue and 5 yellow marbles of the same size. Hari
takes out a marble from the jar at random. What is the probability that the chosen marble is of:
( a) red colour? (b) not a blue colour?

Solution: Total number of marbles in the glass jar = 6 + 5 + 4 + 5 = 20
So, the total number of outcomes = 20

( a) Number of red colour marbles = 6
6 3
\ P(getting a marble of red colour) = =
20 10
( b) Number of blue colour marbles = 4
4 1
\ P(getting a marble of blue colour) = =
20 5
1 4
\ P(getting a marble of not blue colour) = 1 − =
5 5
Practice Time 4E

1. A die is thrown. What is the probability of getting
( a) an odd number? (b) a multiple of 3?

2. A coin is tossed 200 times and the following result is obtained:
Heads = 105 and Tails = 95

Find the probability of obtaining
( a) heads. (b) tails.

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