Question

Two coins are tossed five times, the probability that an odd number of heads are obtained is:
A. ${\left( {\dfrac{1}{2}} \right)^5}$
B. $\dfrac{3}{5}$
C. $\dfrac{2}{5}$
D. $\dfrac{1}{2}$

Answer 1

Hint : This question can be solved using discrete probability distribution.
$P(X = r) = {}^n{C_r}{p^r}{q^{n - r}}$
Where p denotes success,
q denotes failure
n is total trials and
r is the number of successors

→p + q = 1
Here, success = getting head odd number of times.

Complete step-by-step answer :
We have,
Number of trials (n) =5
Number of success [odd number of heads] (r)=1,3,5
Probability of getting heads (p)= $\dfrac{1}{2}$
Probability of getting tails (q)= $\dfrac{1}{2}$
Required probability [P(X=r)]:
$P(X = 1) + P(X = 3) + P(X = 5)$
Substituting the values in probability distribution, we get:
${}^5{C_1}{\left( {\dfrac{1}{2}} \right)^1}{\left( {\dfrac{1}{2}} \right)^4} + {}^5{C_3}{\left( {\dfrac{1}{2}} \right)^3}{\left( {\dfrac{1}{2}} \right)^2} + {}^5{C_5}{\left( {\dfrac{1}{2}} \right)^5}{\left( {\dfrac{1}{2}} \right)^0}$
Simplifying:
$= {\left( {\dfrac{1}{2}} \right)^5}\left( {5 + \dfrac{{5 \times 4}}{2} + 1} \right)$
$= \dfrac{1}{{32}}\left( {\dfrac{{32}}{2}} \right) = \dfrac{{16}}{{32}}$
$= \dfrac{1}{2}$

Therefore, the probability two obtain an odd number of heads when two coins are tossed five times is $\dfrac{1}{2}$ , option D).
So, the correct answer is “Option D”.

Note : Another approach to solve the question can be noting the possible outcomes and calculating the probability.
Possible outcomes: [5H, 5T, (H, 4T),(2H,3T),(T,4H),(3H,2T)]
It can be such that,
Total outcomes=6
Odd number of heads obtained =3
Now, ${\text{probability }} = \dfrac{{{\text{Favourable outcomes}}}}{{{\text{Total outcomes}}}}$
$= \dfrac{3}{6}$
$= \dfrac{1}{2}$
which is the required answer.

This binomial probability distribution can be used when the number of trials are fixed and independent, probability of success or failure is exactly the same for all trials.