Question

The fair coin is tossed simultaneously. If ‘tail’ appears on the first four tosses then find the probability of the head that will be appearing on the fifth toss.
(a)$\dfrac{1}{1}$
(b)$\dfrac{1}{3}$
(c)$\dfrac{1}{2}$
(d)$\dfrac{2}{5}$

Answer 1

Hint: The given problem revolves around the concepts of probability. So, we will use the definition of probability for each iteration/s for a given number of outcomes. The probability of an event A is denoted by ‘P(A)’ and the basic formula of finding probability is $P(A) = \dfrac{{n(s)}}{{n(A)}}$ , where n(s) denotes the favorable outcomes and n(A) denotes the total number of outcomes for an respective event.

Complete step by step answer:
Since, the coin is fair, both the outcomes that are possibilities (‘head’ and ‘tails’) are equally probable; the probability of ‘heads’ is equally authorized with probability of getting ‘tails’ (where no other possibilities are possible in this case particularly).
Hence, probability of getting ‘head’ is $ = \dfrac{1}{2}$ and that of Probability of getting ‘tail’ is $ = \dfrac{1}{2}$ respectively (that it shows the $50\% $ of possibility of both the outcomes or also written as $ = 0.5$).
Where, $\text{Probability} = \dfrac{{{\text{ number of favourable outcomes}}}}{{{\text{total number of outcomes in the event}}}}$ (here, number of outcomes is $1$ and total outcomes in the event is $1$)
Now, we are given that ‘tail’ appears on the first four tosses. We know that the probability of getting a tail on a coin toss is equal to half.
That is, at first iteration, \[P_1 = \dfrac{1}{2}\]
At second iteration, \[P_2 = \dfrac{1}{2}\]
At third iteration, \[P_3 = \dfrac{1}{2}\]
And, at fourth iteration, \[P_4 = \dfrac{1}{2}\]
As a result, we can say that the events of ‘heads’ to that of ‘tail’ are independent of the required outcome.
So, it seems to be similar with the ‘tails’ outcome that the ‘heads’ outcome will also have equal probability at any iteration of the event being sampled (as asked in the question)
$\therefore $ The probability of getting ‘head’ for fifth time is $ = \dfrac{1}{2}$ respectively. The correct option is option (c).

Note:
One should know the basic formula of probability while solving a question. We should also know the concepts of independent events and that their probabilities are not affected by each other’s occurrences. We should take care of the calculations so as to be sure of our final answer. Also, we should know that the sum of probabilities of all the possible events in a situation is one. That’s why, the sum of getting heads and tails on tossing a coin is one.