Question

An unbiased coin is tossed three times. If A is the event that a head appears on each
of the first two tosses, B is the event that a tail occurs on the third toss and C is the
event that exactly two tails appear in the three tosses, show that:
i) Events A and B are independent
ii) Events B and C are dependent.

Answer 1

Answer on Question #62211 – Math – Statistics and Probability

Question

An unbiased coin is tossed three times. If $A$ is the event that a head appears on each of the first two tosses, $B$ is the event that a tail occurs on the third toss and $C$ is the event that exactly two tails appear in the three tosses, show that:

i) Events $A$ and $B$ are independent;

ii) Events $B$ and $C$ are dependent.

Solution

Obviously $P(A) = P\{\text{head}, \text{head}, *\} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$, $P(B) = P\{*, *, \text{tail}\} = \frac{1}{2}$, where * may be either head or tail.

To calculate $P(C) = P\{\text{exactly two tails appear in the three tosses}\}$ we must use the binomial distribution. In our case $n = 3$, $p = \frac{1}{2}$ (the probability of having the tail on one toss),

$$q = 1 - p = \frac{1}{2}. \text{ So } P(C) = C_3^2 \cdot \left(\frac{1}{2}\right)^2 \cdot \left(\frac{1}{2}\right)^{3-2} = \frac{3!}{2! \cdot 1!} \cdot \left(\frac{1}{2}\right)^3 = \frac{3}{8}.$$

i)

$$P(A \cap B) = P\{\text{head}, \text{head}, \text{tail}\} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}.$$

Since $\frac{1}{8} = P(A \cap B) = P(A)P(B) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}$, we conclude that events $A$ and $B$ are independent.

ii)

$$P(B \cap C) = P\{\text{tail}, \text{head}, \text{tail}\} + P\{\text{head}, \text{tail}, \text{tail}\} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.$$

Since $\frac{1}{4} = P(B \cap C) \neq P(B)P(C) = \frac{1}{2} \cdot \frac{3}{8} = \frac{3}{16}$, we conclude that events $B$ and $C$ are dependent.

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