Question

(a)Two unbiased coins are tossed simultaneously. Find the probability of getting
(i)Exactly one head
(ii)Exactly one tail
(iii)Two tails
(iv)At least one tail
(v)At most one tail
(vi)No tail

(b)Harpreet tossed two different coins simultaneously. What is the probability that she gets at least one head?

Answer 1

Hint: In order to find the sub parts of the statements, we need to find the sample space or total number of favourable outcomes. Then using the sample space we need to find the number of favourable outcomes, then substituting all the values in the formula to find the probability of an event to get the value.

Complete step by step solution:
The formula of finding probability is:
$P\left( A \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total Number of favourable outcomes}}}}$
(a)We are given that two unbiased coins are tossed simultaneously. So, the sample space of the condition is written as:
Sample Space(S) $ = \left\{ {\left( {H,H} \right),\left( {H,T} \right),\left( {T,H} \right),\left( {T,T} \right)} \right\}$
Total number of favourable outcomes is 4 (as counted from the above sample space).

(i) Exactly one head- In the Sample we can see that there are two values that contains exactly one head, that are: $\left( {H,T} \right),\left( {T,H} \right)$
So, Number of favourable outcomes $ = 2$
Substituting the values in formula of probability:
So, the probability will be:
$P\left( {{\text{Exactly one head}}} \right) = \dfrac{2}{4} = \dfrac{1}{2}$

(ii) Exactly one tail- In the Sample we can see that there are two values that contains exactly one tail, that are: $\left( {H,T} \right),\left( {T,H} \right)$
So, Number of favourable outcomes $ = 2$
Substituting the values in formula of probability:
So, the probability will be:
$P\left( {{\text{Exactly one tail}}} \right) = \dfrac{2}{4} = \dfrac{1}{2}$

(iii) Two Tails- In the Sample we can see that there is only one value that contains two tails, that is: $\left( {T,T} \right)$
So, Number of favourable outcomes$ = 1$
Substituting the values in formula of probability:
So, the probability will be:
$P\left( {{\text{Two tails}}} \right) = \dfrac{1}{4}$

(iv) At least one tail- Since, it is given at least one tail that means there can be more than one tail. So, In the Sample we can see that there are three values that contains at least one tail, that are: $\left( {H,T} \right),\left( {T,H} \right),\left( {T,T} \right)$
So, Number of favourable outcomes$ = 3$
Substituting the values in formula of probability:
So, the probability will be:
$P\left( {{\text{At least one tail}}} \right) = \dfrac{3}{4}$

(v) At most one tail- Since, it is given at most one tail that means there can be one tail for maximum that means it can also have no tail. So, In the Sample we can see that there are three values that contains at most one tail, that are: $\left( {H,T} \right),\left( {T,H} \right),\left( {H,H} \right)$
So, Number of favourable outcomes $ = 3$
Substituting the values in formula of probability:
So, the probability will be:
\[P\left( {{\text{At most one tail}}} \right) = \dfrac{3}{4}\]

(vi) No tail- In the Sample we can see that there is only one value that contains no tail, that is: $\left( {H,H} \right)$
So, Number of favourable outcomes $ = 1$
Substituting the values in formula of probability:
So, the probability will be:
$P\left( {{\text{No tail}}} \right) = \dfrac{1}{4}$

(b)Harpreet tossed two different coins simultaneously, so the sample space of the condition would be:
Sample Space(S) $ = \left\{ {\left( {H,H} \right),\left( {H,T} \right),\left( {T,H} \right),\left( {T,T} \right)} \right\}$
Total number of favourable outcomes is 4 (as counted from the above sample space).
Now, for getting at least one head- Since, it is given at least one head that means there can be more than one head. So, In the Sample we can see that there are three values that contains at least one head, that are: $\left( {H,T} \right),\left( {T,H} \right),\left( {H,H} \right)$
So, Number of favourable outcomes$ = 3$
Substituting the values in formula of probability:
So, the probability will be:
$P\left( {{\text{At least one head}}} \right) = \dfrac{3}{4}$


Note:
Unbiased coins means that the values on both the sides are different and it’s a fair coin, and vice versa would be that the value on both the sides of the coins would be same, either head and head or tail and tail. Biased coin will be an unfair coin.