Question

Three coins are tossed together, find the probability of
getting: i) exactly two heads ii) at least two heads
iii) at most two heads.​

Answer 1

Step-by-step explanation:

When three coins are tossed together, the total number of outcomes =8

i.e., (HHH,HHT,HTH,THH,TTH,THT,HTT,TTT)

Solution (i):

Let E be the event of getting exactly two heads

Therefore, no. of favorable events, n(E)=3(i.e.,HHT,HTH,THH)

We know that, P(E) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

8

3

Solution (ii):

Let F be the event of getting atmost two heads

Therefore, no. of favorable events, n(E)=7(i.e.,HHT,HTH,TTT,THH,TTH,THT,HTT)

We know that, P(F) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

8

7

Solution (iii):

Let H be the event of getting at least one head and one tail

Therefore, no. of favorable events, n(H)=6(i.e.,HHT,HTH,THH,TTH,THT,HTT)

We know that, P(H) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

8

6

=

4

3

Solution (iv):

Let I be the event of getting no tails

Therefore, no. of favorable events, n(I)=1(i.e.,HHH)

We know that, P(H) =

(Total no.of possible outcomes)

(No.of favorable outcomes)

=

8

1