Three coins are tossed simultaneously for 1000 times and outcomes are noted

Find the probability of getting
(i) 2 head
(ii) More than 1 head

HINT- In order to solve such a type of question we must proceed through finding favourable cases and total cases, then by applying formulas of probability we can reach our answer.

Complete step-by-step answer:

In the question, it is given that Three coins are tossed simultaneously for 1000 times hence total cases will be 1000
i.e. total outcomes =1000

(i) for 2 head

As given in question, two heads occurred for 200 times
So number of favourable outcomes = 200
total number of possible cases = 1000

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (two heads) =$\dfrac{{200}}{{1000}} = \dfrac{1}{5}$

ii) For more than 1 head
For more than 1 head there are two case either 2 head or 3 head
we know that
In the case of OR, total cases got added

As given in question, two heads occurred for 200 times and three heads occurred for 30 times
So number of favourable outcomes = 200+30=230
total number of possible cases = 1000

As we know that Probability \[\left( {\text{P}} \right){\text{ = }}\dfrac{{{\text{Favourable number of cases}}}}{{{\text{Total number of cases}}}}\]
Probability (two heads) =$\dfrac{{230}}{{1000}} = \dfrac{{23}}{{100}} = 0.23$

Note- In Such Types of Question where number of tossed are more, we can’t calculate total outcomes that we generally used to do when three coins are tossed once or twice.
So where the number of tossed are more either total outcomes will be given in the question or we have to go for Poisson distribution which is part of engineering mathematics.