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A coin is biased so that a head is twice as likely to occur as a tail. If the coin is tossed 3 times, then the probability of getting two tails and one head is

JEE · Math · previous-year question

  1. A.$$\frac{1}{9}$$
  2. B.$$\frac{2}{9}$$correct
  3. C.$$\frac{1}{27}$$
  4. D.$$\frac{2}{27}$$

Answer

B. $$\frac{2}{9}$$

Explanation

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

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