Question

Example 35. Three coins are tossed simultaneously. Find the probability of getting :
(1) three heads (iii) one head
(iv) no head
(v) atleast one head
(vi) atleast two heads
(vil) atmost two heads
(CBSE 2015)
(CBSE 2013)​

Answer 1

Answer:

Solution (i): 3

Solution (i): 3 ⎯

⎯ 8

8Solution (ii): 7

8Solution (ii): 7 ⎯

⎯ 8

8Solution (iii): 3

8Solution (iii): 3 ⎯

⎯ 4

4Solution (iv): 1

4Solution (iv): 1 —

— 8

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) 3

(Total no.of possible outcomes) 3⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = ⎯

⎯ (No.of favorable outcomes) 8

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) 7

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = ⎯

(No.of favorable outcomes) 8

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) 6 3

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = ⎯ = ⎯

(No.of favorable outcomes) 8 4

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) 1

Total no.of possible outcomes) 1⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ = ⎯

= ⎯ (No.of favorable outcomes) 8

⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

Hope This Helpful

Thankyou