Question

A coin is tossed twice. What is the probability of getting two consecutive tails?
$(a){\text{ }}\dfrac{1}{2} \\\ (b){\text{ }}\dfrac{1}{4} \\\ (c){\text{ }}\dfrac{1}{8} \\\ (d){\text{ None of these}} \\\$

Answer 1

Hint – In this question there can be two ways to approach the problem, we know that the probability of getting one tail is $\dfrac{1}{2}$, using basic probability definition so use this to obtain probability of two consecutive that is one after another tail. Another approach we will discuss in the later end.

Complete step-by-step answer:

As we know a coin has two possibilities either head or tail.

So total number of outcomes = 2

And we know that the probability (P) is the ratio of favorable number of outcomes to the total number of outcomes.

So the probability (P) of getting a tail when a coin is tossed is
$\Rightarrow P = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}} = \dfrac{1}{2}$.

Now it is given that the coin is tossed twice so the probability (P1) of getting two consecutive tails is
$\Rightarrow {P_1} = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$

So this is the required probability of getting two consecutive tails.

Hence option (B) is correct.

Note – Another approach is by considering all the possible cases that may appear while throwing two coins simultaneously, and the favorable cases of getting two consecutive tails will be one only that is$\left( {T,T} \right)$. Use the basic probability formula to get the answer. Key point is in this question the coin is unbiased that means it’s not weighted towards one side thus the probability of head and tails are both equal to $\dfrac{1}{2}$, if it would have been weighed more towards one side than we can say both of these probabilities to be equal.