Question

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:

Outcome:	No head	One head	Two heads	Three heads
Frequency:	14	38	36	12

If the three coins are simultaneously tossed again, compute the probability of:
(i) 2 heads coming up.
(ii) 3 heads coming up.
(iii) at least one head coming up.
(iv) getting more heads than tails.
(v) getting more tails than heads.

Answer 1

The total number of trials is 100.

Remember the empirical or experimental or observed frequency approach to probability.

If n be the total number of trials of an experiment and A is an event associated to it such that A happens in m-trials. Then the empirical probability of happening of event A is denoted by P (A ) and is given by

P (A) =`m/n` 

(i) Let A be the event of getting two heads.

The number of times A happens is 36.

Therefore, we have

 P (A) =`36 /100` 

=0.36

(ii) Let B be the event of getting three heads

The number of times B happens is 12.

Therefore, we have

 P (B) =`12/100` 

=0.12

(iii) Let C be the event of getting at least one head.

The number of times C happens is 38+36+12=86.

Therefore, we have

P (c) =`86/100` 

=0.86

(iv) Let D be the event of getting more heads than tails.

The number of times D happens is 36+ 12+ 48 .

Therefore, we have

P (D) =`48/100` 

=0.48

(v) Let E be the event of getting more tails than heads.

The number of times E happens is 14+38+=52.

Therefore, we have

P (E) =`52/100` 

=0.52 .