Question

If a coin is tossed 3 times the probability of obtaining 2 heads or 2 tails is
A) $\dfrac{1}{4}$
B) $\dfrac{1}{2}$
C) $\dfrac{2}{3}$
D) $\dfrac{3}{4}$

Answer 1

Hint:
It is given in the question that If a coin is tossed 3 times and we have to find the probability of obtaining 2 heads or 2 tails.
First, we will find the total favorable outcomes. Then after we will find the total favorable outcomes of 2 heads and 2 tails.
Finally, we will find the probability of 2 heads or 2 tails and we will get the answer.

Complete step by step solution:
It is given in the question that If a coin is tossed 3 times the probability of obtaining 2 heads or 2 tails is
If a coin is tossed three times then the total favorable outcomes \[ = \left( {H,H,H} \right),\left( {T,T,T} \right),\left( {H,H,T} \right),\left( {H,T,H} \right),\left( {T,H,H} \right),\left( {T,T,H} \right),\left( {T,H,T} \right),\left( {H,T,T} \right) = 8\]
All are heads \[ = \left( {H,H,H} \right) = 1\] .
All are tails \[ = \left( {T,T,T} \right) = 1\] .
2 heads \[ = \left( {H,H,T} \right),\left( {H,T,H} \right),\left( {T,H,H} \right) = 3\]
2 tails \[ = \left( {T,T,H} \right),\left( {T,H,T} \right),\left( {H,T,T} \right) = 3\]
 2 heads and 2 tails i.e. \[\left( {H,H,T} \right),\left( {H,T,H} \right),\left( {T,H,H} \right),\left( {T,T,H} \right),\left( {T,H,T} \right),\left( {H,T,T} \right) = 6\]

$\therefore $ Probability of 2 heads and 2 tails $ = \dfrac{6}{8} = \dfrac{3}{4}$.

Note:
Probability: Probability is the number of ways of achieving success, the total number of possible outcomes.
For Example: The probability of flipping a coin and its being head is $\dfrac{1}{2}$ because there is 1 way of getting a head and the total number of possible outcomes is 2 (head or tail). We write P(head) = $\dfrac{1}{2}$