QJEEM24Ma3b8a: JEE Main 2024 Mathematics - Class 12 JEEM24Ma3b8a Solution: A coin is biased so that a head is twice as likely to occur as…

To solve this problem, we need to first determine the probability of getting a head (H) and the probability of getting a tail (T).

Since a head is twice as likely to occur as a tail, we can denote the probability of getting a tail as $P(T) = p$ and the probability of getting a head as $P(H) = 2p$ .

These probabilities must sum to 1 because those are the only two possible outcomes for each coin toss :

$P(H) + P(T) = 1$

$2p + p = 1$

$3p = 1$

$p = \frac{1}{3}$

Therefore, the probability of getting a tail (T) is $P(T) = \frac{1}{3}$ and the probability of getting a head (H) is $P(H) = 2 \times \frac{1}{3} = \frac{2}{3}$ .

Now to find the probability of getting two tails and one head, we need to consider the different sequences in which this can occur. There are three unique sequences: TTH, THT, and HTT.

The probability of each sequence is found by multiplying the probabilities of each individual event since each coin toss is independent:

$P(TTH) = P(T) \times P(T) \times P(H) = \left(\frac{1}{3}\right)^2 \times \frac{2}{3} = \frac{1}{9} \times \frac{2}{3} = \frac{2}{27}$

$P(THT) = P(T) \times P(H) \times P(T) = \frac{1}{3} \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{27}$

$P(HTT) = P(H) \times P(T) \times P(T) = \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{2}{27}$

The overall probability of getting two tails and one head in any order is the sum of these individual probabilities :

$P(2T1H) = P(TTH) + P(THT) + P(HTT) = \frac{2}{27} + \frac{2}{27} + \frac{2}{27} = \frac{6}{27}$

Simplifying this expression gives us: $P(2T1H) = \frac{6}{27} = \frac{2}{9}$

Therefore, the correct answer is :

Option B : $\frac{2}{9}$