Question

Consider the experiment of tossing a coin. If the coin shows head, toss it again; but if it shows a tail, then throw a die. Find the conditional probability of the event A: `the die shows a number greater than 3' given that B: `there is at least one tail'.

Hint:

When working with conditional probabilities, remember the formula $P(A|B) = \frac{P(A \cap B)}{P}$. Also, when calculating probabilities from distributions, ensure that the total probability adds up to 1 and check for consistency in calculations.

Answer 1

Step 1: List the possible outcomes of the experiment

Sample space:

• HH → coin tossed twice, no tail → not included in B
• HT → second toss is tail → throw die → outcomes: HT1, HT2, HT3, HT4, HT5, HT6
• TH → first toss is tail → throw die → outcomes: TH1, TH2, TH3, TH4, TH5, TH6
• TT → both tosses tails → throw die → outcomes: TT1, TT2, TT3, TT4, TT5, TT6

Total possible outcomes = HH, HT1–HT6, TH1–TH6, TT1–TT6 → total = 1 + 18 = 19 outcomes

Step 2: Define Event B: At least one tail

All outcomes except HH have at least one tail.
So, total favorable outcomes for B = 18

Step 3: Define Event A: die shows number > 3

This means die shows 4, 5, or 6 → for outcomes involving die:

• HT4, HT5, HT6
• TH4, TH5, TH6
• TT4, TT5, TT6

Total = 9 outcomes

Step 4: Find A ∩ B (both A and B happen)

Outcomes where there is at least one tail and die shows > 3 = the 9 outcomes listed above.
So, $n(A \cap B) = 9$

Step 5: Conditional Probability:

$\displaystyle P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{9}{18} = \mathbf{\dfrac{1}{2}}$

Final Answer:

$\boxed{ \dfrac{1}{2} }$