Question

Three coins are tossed simultaneously find the probability of getting
a) $3$ heads
b) $2\,heads$
c) no heads
d) at least 1 head
e) at most 2 heads

Answer 1

Hint : For this problem we first find sample space for three coins and then look for favorable cases according to the given part to its respective probability.
Probability of any event = $\dfrac{{Number\,\,of\,\,favourable\,\,cases}}{{Total\,\,number\,\,of\,\,sample\,spaces}}$

Complete step-by-step answer :
For probability problems the first step of every question is to write its sample space.
Here, three coins are tossed together.
Therefore sample spaces are given as:
(H, H, H), (T, T, T)
(H, H, T), (H, T, H), (T, H, H)
(T, T, H), (T, H, T), (H, T, T)
Hence, for three coins there are $8$ sample spaces or we can say there are $8$ outcomes.
Now we will discuss probability of each given case one by one
a) $3$ Heads
To find probability we will first find favorable out of given sample spaces.
For $3$ heads there is only one case in favor which is (H, H, H)
Hence, its probability will be given as = $\dfrac{{No.\,of\,favorable\,cases}}{number\,\,of\,elements\,\,in\,\,Sample\,spaces}$
$\Rightarrow$ Probability of three heads = $\dfrac{1}{8}$

b) $2\,heads$
In case of $2\,heads$ number of favorable cases are (H, H, T), (H, T, H), (T, H, H) = $3$
Hence, its probability is given as = $\dfrac{{No.\,of\,favorable\,cases}}{{number\,\,of\,elements\,\,in\,\,Sample\,spaces}}$
$\Rightarrow$ Probability of two heads = $\dfrac{3}{8}$

c) no heads
In this case number of favorable cases are (T, T, T) = $1$
Hence, its probability is given as = $\dfrac{{No.\,of\,favorable\,cases}}{{number\,\,of\,elements\,\,in\,\,Sample\,spaces}}$
$\Rightarrow$ Probability of no heads = $\dfrac{1}{8}$

d) at least one head
In this case we do consider one head, two heads and three heads as favorable cases.
Therefore, total number of favorable cases are (H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T) = $7$
Hence, its probability is given as = $\dfrac{{No.\,of\,favorable\,cases}}{number\,\,of\,elements\,\,in\,\,Sample\,spaces}$
$\Rightarrow$ Probability of at least one head = $\dfrac{7}{8}$

e) at most $2$ heads
IN this case we do consider all cases having either $2$ heads or less than two.
Therefore, total number of favorable cases are (T, T, T), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T) = $7$
Hence, its probability is given as = $\dfrac{{No.\,of\,favorable\,cases}}{number\,\,of\,elements\,\,in\,\,Sample\,spaces}$
$\Rightarrow$ Probability of at most two heads = $\dfrac{7}{8}$
Which is the required simplification for different parts of probability in toss of three coins.

Note : In case of probability problems should make sample space of the given problem very carefully if they make wrong or less sample space then the whole calculation will be wrong and no use of doing such calculations. Therefore, after reading the statement double check sample space formed.