Question

A coin is tossed thrice. The probability of getting a head on the second toss given that a tail has occurred in at least two tosses is 

• A. \(\dfrac{1}{2}\)
• B. \(\dfrac{1}{16}\)
• C. \(\dfrac{1}{8}\)
• D. \(\dfrac{1}{4}\)
• E. \(\dfrac{1}{3}\)

Answer 1

The Correct Option is D

Let \( H \) represent heads and \( T \) represent tails.

The sample space for tossing a coin three times is:

\[ S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\} \]

There are 8 equally likely outcomes.

Let \( A \) be the event that a tail occurs in at least two tosses.

\[ A = \{HTT, THT, TTH, TTT\} \]

There are 4 outcomes in \( A \).

\[ P(A) = \frac{4}{8} = \frac{1}{2} \]

Let \( B \) be the event that a head occurs on the second toss.

\[ B = \{HTH, HTT, THH, THT\} \]

There are 4 outcomes in \( B \).

\[ P(B) = \frac{4}{8} = \frac{1}{2} \]

We want to find the probability of getting a head on the second toss given that a tail has occurred in at least two tosses. This is \( P(B|A) \).

Using the conditional probability formula:

\[ P(B|A) = \frac{P(B \cap A)}{P(A)} \]

\( B \cap A \) is the event that a head occurs on the second toss and a tail occurs in at least two tosses.

\[ B \cap A = \{HTT, THT\} \]

There are 2 outcomes in \( B \cap A \).

\[ P(B \cap A) = \frac{2}{8} = \frac{1}{4} \]

Therefore,

\[ P(B|A) = \frac{P(B \cap A)}{P(A)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2} \]

The probability of getting a head on the second toss given that a tail has occurred in at least two tosses is \( \frac{1}{2} \).

Answer 2

Let’s define the events more precisely:

• A: At least two tails occur. This includes the outcomes {HTT, THT, TTH, TTT}. Thus, \( P(A) = \frac{4}{8} = \frac{1}{2} \).
• B: A head occurs on the second toss. This includes the outcomes {HTH, HTT, THH, THT}. Thus, \( P(B) = \frac{4}{8} = \frac{1}{2} \).

The intersection \( A \cap B \) is the event where both conditions hold: at least two tails occur and there’s a head on the second toss. The outcomes satisfying this are {HTT, THT}. 

Therefore:

\[ P(A \cap B) = \frac{2}{8} = \frac{1}{4}. \]

Now, we can calculate the conditional probability \( P(B|A) \):

\[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}. \]