Question

Three coins are tossed once. Find the probability of getting
(i) 3 heads
(ii) At least 2 heads
(iii) At most two tails
(iv) No tail

Answer 1

Hint:We solve this problem first by using the simple formula of probability. If E be the event, then the probability of occurring the event E is given as
\[\Rightarrow P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}=\dfrac{N}{T}\]
By using this formula we solve all the parts we are asked to find.

Complete step-by-step solution:
We are given that 3 coins are tossed simultaneously
We know that each toss has 2 possibilities either head or tail and it was tossed three times
So, we can say that the total number of outcomes is
\[\Rightarrow T={{2}^{3}}=8\]
Now let us solve the parts one by one.
(i) Getting 3 heads
We know that there will be only one possibility which is getting all 3 heads
So, the possible outcomes are
\[\Rightarrow N=1\]
We know that the probability formula is given as
\[\Rightarrow P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}=\dfrac{P}{T}\]
By using this formula we get the probability of getting 3 heads as
\[\Rightarrow P=\dfrac{1}{8}\]
(ii) Getting at least 2 heads
We know that at least 2 heads mean the total possible outcomes will be the sum of getting 2 heads and getting 3 heads that is
\[\Rightarrow N=\text{Getting 2 heads}+\text{getting 3 heads}\]
We know that the possible outcomes of getting 2 heads are \[{}^{3}{{C}_{2}}\] by substituting this number in the above equation we get
\[\begin{align}
  & \Rightarrow N={}^{3}{{C}_{2}}+1 \\
 & \Rightarrow N=3+1=4 \\
\end{align}\]
We know that the probability formula is given as
\[\Rightarrow P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}=\dfrac{P}{T}\]
By using this formula we get the probability of getting at least 2 heads as
\[\Rightarrow P=\dfrac{4}{8}=\dfrac{1}{2}\]
(iii) Getting at most two tails
We know that the number of possible outcomes of getting at most 2 tails is the sum of getting 2 tails, getting 1 tail, and getting no tail
By using the above relation we get the possible outcomes as
\[\begin{align}
  & \Rightarrow N={}^{3}{{C}_{2}}+{}^{3}{{C}_{1}}+{}^{3}{{C}_{0}} \\
 & \Rightarrow N=3+3+1=7 \\
\end{align}\]
We know that the probability formula is given as
\[\Rightarrow P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}=\dfrac{P}{T}\]
By using this formula we get the probability of getting at most 2 tails as
\[\Rightarrow P=\dfrac{7}{8}\]
(iv) Getting no tail
We know that there will be only one possibility which is getting no tail that is getting all 3 heads
So, the possible outcomes are
\[\Rightarrow N=1\]
We know that the probability formula is given as
\[\Rightarrow P\left( E \right)=\dfrac{\text{number of possible outcomes}}{\text{total number of outcomes}}=\dfrac{P}{T}\]
By using this formula we get the probability of getting no tail as
\[\Rightarrow P=\dfrac{1}{8}\]

Note: This problem can be done in another method to find the possible outcomes.
We can write all the possible outcomes as follows
${ (H, H, H), (H, H, T), (H, T, H), (T, H, H), (H, T, T), (T, H, T), (T, T, H)} $
Here we get all favorable outcomes by counting them for all given conditions like for no tails means all are head{H, H, H}.