JEE Main 2024 — Probability Question with Solution

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