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Question 1:

Toss a coin 4 times. Let $A$ denote the event that a head is obtained on the first toss, and let $B$ denote the event that a head is obtained on the fourth toss. Is $A \cap B$ empty?

I'm not sure how to understand $A \cap B$. Does it mean that 'After tossing 4 times, the first and the fourth time is head'?

If so, can I calculate $P(A \cap B) = {1 \over 2}^2 = {1\over4}$, because for each time, $P(\text{head}) = {1 \over 2}$.

I'm also confused with the formula $P(A|B)$. Should I use this formula here? Whether or not, tell me why, please.

Question 2:

A well-shuffled 52-card deck is dealt to 4 players. Find the probability that each of the player get an Ace.

How to approach this problem? I totally have no idea.

Thanks in advance :P

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    http://math.stackexchange.com/questions/1345413/is-there-an-alternative-intuition-for-solving-the-probability-of-having-one-ace –  Jul 01 '15 at 08:21

1 Answers1

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Question 1:

Your understanding of $A\cap B$ and your calculation of $P(A\cap B)$ are correct.

You basically did use $P(A\mid B)$, intuitively. To expand your calculation, since $A$ and $B$ are clearly independent events by the nature of coin tossing, $P(A\mid B)=P(A)$. Therefore

$$P(A\cap B)=P(A\mid B)\cdot P(B)=P(A)\cdot P(B)=\frac 12\cdot \frac 12= \left(\frac 12\right)^2=\frac 14$$

You did not think through each step there, but you still did the correct calculation.

Question 2:

I can't improve on @d.k.o.'s suggestion to see this question with its answers. I especially liked @joriki's answer.

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Rory Daulton
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