Question

three coins are tossed once.find probability of getting (i) atleast 2 heads (ii) atmost 2 tails​

Answer 1

Step-by-step explanation:

is it correct? kindly reply for it.

Answer 2

Step-by-step explanation:

ANSWER

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)=83

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)=87

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)=86=43

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)=81