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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.
Solution
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 .
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