Question

A coin is tossed four times, if H = head and T = tail, what is the probability of the tosses coming up in the order HTHH.
A) $\dfrac{3}{{16}}$
B) $\dfrac{1}{{16}}$
C) $\dfrac{5}{{16}}$
D) $\dfrac{7}{{16}}$

Answer 1

Hint:
Firstly, find the number of elements in the Universal set of the above question.
Then, find the number of elements that are similar to the event of getting HTHH.
Thus, find the probability of getting HTHH by dividing the number of terms in event of getting HTHH by the number of elements in the Universal set.

Complete step by step solution:
It is given that a coin is tossed 4 times.
Thus, the universal set of the event that coin is tossed 4 times will be U = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THTH, HTTT, THTT, TTHT, TTTH, TTHH, HTHT, THHT, TTTT}
Thus, \[n\left( U \right) = 16\].
Now, it is asked to find the probability of the tosses coming up in the order HTHH.
Let X be the event that the tosses come in the order of HTHH.
In the above-mentioned Universal set, the event that the coin tossed 4 times shows HTHH occurs one time only. Thus, \[n\left( X \right) = 1\] .
Now, probability of event x is given by dividing the number of terms in event X by the number of elements in the Universal set.
Thus, $P(X) = \dfrac{{n(X)}}{{n(U)}}$
$ = \dfrac{1}{{16}}$

So, option (B) is correct.

Note:
The shortcut method to find the number of elements in the Universal set of the event that a coin is tossed n times is as follows:
The number of elements in the Universal set of the event that a coin is tossed n times is given by $n\left( U \right) = {2^n}$.
Here, the coin is tossed 4 times. So, $n\left( U \right) = {2^4} = 16$.