Question

In a simultaneous toss of four coins, what is the probability of getting exactly three heads?
A. $\dfrac{1}{2}$
B. $\dfrac{1}{3}$
C. $\dfrac{1}{4}$
D. None of these

Answer 1

In the given question, we have to find the probability of getting exactly three heads in a total of three tosses. So, we will make use of the multiplication rule of probability to find the probability of two events happening one after another. We will first find the probability of getting a single head on a toss. Then, we will make use of the binomial probability distribution formula to get to the required answer.

Complete step by step answer:
So, we have the number of coin tosses as $4$. Number of heads required is $3$. When a coin is tossed, the total number of possibilities is $2$. Number of favorable possibilities for heads is $1$. Hence, the probability of getting a head in a coin toss is $\dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}} = \dfrac{1}{2}$.

Now, we have to find the probability of exactly three heads in four tosses. We know that we can choose three out of four tosses in $^4{C_3}$ ways. Also, the probability of getting head in a toss is $\dfrac{1}{2}$. Probability of getting a tail in a toss is $\dfrac{1}{2}$.So, we get the required probability getting two heads out of three tosses as $\left( {^4{C_3}} \right) \times {\left( {\dfrac{1}{2}} \right)^3} \times \left( {\dfrac{1}{2}} \right)$ using the binomial probability distribution.

So, we get, required probability $= \left( {^4{C_3}} \right) \times {\left( {\dfrac{1}{2}} \right)^3} \times \left( {\dfrac{1}{2}} \right)$
We know the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. So, we get,
$4 \times \left( {\dfrac{1}{8}} \right) \times \left( {\dfrac{1}{2}} \right)$
Computing the product, we get,
$\dfrac{1}{4}$

So, we get the probability of getting exactly three heads in four coin tosses is $\left( {\dfrac{1}{4}} \right)$.

Hence, the correct answer is the option C.

Note: We must know the multiplication rule of probability for finding probability of the events happening one after another. One should know the applications of simplification rules to simplify the calculations. We should remember the combination formula $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ to evaluate the number of ways of choosing two out of five attempts.