Question

If three coins are tossed simultaneously then the probability of getting at least two heads is
$
  (a){\text{ }}\dfrac{1}{4} \\
  (b){\text{ }}\dfrac{3}{8} \\
  (c){\text{ }}\dfrac{1}{2} \\
  (d){\text{ }}\dfrac{1}{3} \\
 $

Answer 1

Hint – In this question the probability of getting a single head is $\dfrac{1}{2}$. So if we let X be the event of getting a head then we have to solve for $P(X \geqslant 2)$ but this can go only up to X= 3 as if we toss 3 coins then only 3 heads can show up.

Complete step-by-step solution-
As we know that in a coin there are two faces so the total number of outcomes = 2.
And favorable number of outcomes i.e. head = 1.
So the probability (P) of getting the head is the ratio of favorable number of outcomes to the total number of outcomes.
$ \Rightarrow P = \dfrac{1}{2}$
Now as we know that the total probability is 1.
So the probability of not getting the head $(q) = 1- \dfrac{1}{2} = \dfrac{1}{2}$.
Now the coin is tossed three times so the probability of getting head at least 2 times is
= getting head 2 times + getting head 3 times.
Here we use the formula of getting r times from n times = ${}^n{C_r}{\left( p \right)^r}{\left( q \right)^{n - r}}$
Therefore, the probability of getting at least 2 heads is
$ \Rightarrow {}^3{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{3 - 2}} + {}^3{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{3 - 3}}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property in above equation we have,
$ \Rightarrow {}^3{C_2}{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{3 - 2}} + {}^3{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^{3 - 3}} = \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}}\left( {\dfrac{1}{4}} \right)\left( {\dfrac{1}{2}} \right) + \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}}\left( {\dfrac{1}{8}} \right){\left( {\dfrac{1}{2}} \right)^0}$
Now simplify the above equation and we all know something raise to the power zero is one so we have,
$ \Rightarrow \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}}\left( {\dfrac{1}{4}} \right)\left( {\dfrac{1}{2}} \right) + \dfrac{{3!}}{{3!\left( {3 - 3} \right)!}}\left( {\dfrac{1}{8}} \right){\left( {\dfrac{1}{2}} \right)^0} = 3\left( {\dfrac{1}{8}} \right) + 1\left( {\dfrac{1}{8}} \right)1$
$ \Rightarrow \dfrac{3}{8} + \dfrac{1}{8} = \dfrac{4}{8} = \dfrac{1}{2}$
So this is the required answer.
Hence option (C) is correct.

Note – The probability distribution of a discrete random variable is a list of probabilities associated with each of its possible values. It is also sometimes called the probability function or the probability mass function. A discrete random variable is a variable that represents numbers found by counting, for example the number of marbles in a jar.