The question is

You have a fair and a biased coin. You toss the fair coin once. If it shows head you toss the fair coin again one time. If it shows tail you toss the biased coin one time.

a) assuming the probability of the biased coin showing head is 1/3, find the the probability of the event where the tosses produce different results (Event A)?

So the sample space is S={HH,HT,TH_b,TT_B}

H_b= head of biased coin and T_b= tail of biased coin


The fact that its a biased coin i have no clue how to approach this


2 Answers 2


In general, the sample space is not unique, so you pick one that is convenient.

Presumably the coin tosses are independent.

I would start with something like $\{ T_f T_b, T_f H_b, H_f T_f, H_f H_f \}$.

Then $A = \{ T_f H_b, H_f T_f \}$.

Note that $P[ A ] = P[ \{ T_f H_b\}] + P [\{H_f T_f \} ] $. (This is usually written as $P[T_f H_b] + P[ H_f T_f]$ .)

Since the tosses are independent, we have $P[T_f H_b] = P [T_f] P[H_b] = {1 \over 2} {1 \over 3}$ and the other is computed similarly.


Hint... a probability tree often helps to understand the problem.

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