The result is either (H) or (T). Define the events

  1. D: ”The fourth toss is tails” and
  2. F: ”Heads and tails alternate”.

Additionally, determine: $$D ∩ F$$

Now for $P (D)$ there are 8 possibilities so then: $$P (D) = 1/(2^8) $$ but my friend said that it needs to be: $$8/(2^8) $$ Is that right? If so, why?

Edit:tried to make it more clear. I already know how to get P (F) and $$D ∩ F$$. My main confusion is just with P (D)

  • 1
    $\begingroup$ $P(D)=0.5$ (only look at the $4$-th toss) and $P(D\cap F)=2^{-4}$ (only HTHT is possible). $\endgroup$ – drhab Mar 21 '18 at 9:38
  • $\begingroup$ It's really unclear what you are asking. $P(D)$ or $P(D \cap F)$? Or something else? $\endgroup$ – Harto Saarinen Mar 21 '18 at 9:39
  • $\begingroup$ @drhab but why is that the only one? Can't you also get (H H H T), (HTTT), (HHTT) and etc? $\endgroup$ – gigantea8 Mar 21 '18 at 9:40
  • $\begingroup$ @HartoSaarinen I am asking for P (D) $\endgroup$ – gigantea8 Mar 21 '18 at 9:40
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    $\begingroup$ If you are talking about $D$ (fourth toss a tail) think of it like this: there are two equiprobable possibilities (head or tail) and their sum is $1$, so both equalize $0.5$. The $4$-th toss has nothing to do with the other tosses (independence). They have no influence at all on what happens at the $4$-th toss. So it is enough to focus on the $4$-th toss. $\endgroup$ – drhab Mar 21 '18 at 9:54

$P(D)$ is $1/2$
Now let us see how...
By Laplace's law, Probability can be expressed as the ratio of the number of favourable events to the total number of events.
Now if you know how to count permutations, you will see that there are clearly $2×2×2×1$ favourable events and $2×2×2×2$ total number of events. Taking the ratio, you get $1/2$

Let us now find $P(F)$

For F, there are two favourable cases, namely, HTHT,THTH. And the total cases remain same.
So, $P(F)=1/2^3$

Finally, $$P(D \cap F)=P(D)×P(F)=1/2^4$$

EDIT: Another way of finding $P(D)$
Clearly $P(D)=1/2$ because what we get on coins 1,2 and 3 has no effect on what we get on 4(independant events) and also there are no restrictions on the outcome of 1,2,3.

2nd EDIT Let us define four events:.
1-Getting either a head or a tail on tossing coin 1.
2-Getting either a head or a tail on tossing coin 2.
3.Getting either a head or a tail on tossing coin 3.
4. Getting a tail on tossing coin 4.
NB All these events are independent. We are not tossing the other coins when we are tossing some $i$th coin.
Therefore, clearly $P(D)=1•1•1•1/2=1/2$

  • 2
    $\begingroup$ I would suggest a minor improvement: The ratio of the number of favourable events to the total number of events is not the definition of probability. It's Laplace's law in the case that all outcomes are equally likely. $\endgroup$ – Philipp Imhof Mar 21 '18 at 10:04

First, $P(D)=\frac12$, because other tosses do not matter.

For $P(F)$, we see that there are $2^4$ possible outcomes when tossing a coin 4 times. Of these, only 2 are good: THTH and HTHT. This makes a total of $P(F)=\frac{2}{2^4}=\frac{1}{2^3}=\frac{1}{8}$.

Now we combine $D$ and $F$. Then, only HTHT is left. So we have $P(D\cap F)=\frac{1}{2^4}=\frac{1}{16}$.

  • 1
    $\begingroup$ The coin is tossed 4 times. $\endgroup$ – Harto Saarinen Mar 21 '18 at 9:41
  • $\begingroup$ OMG, thanks.... I'll edit my answer. I somehow got stuck on that 8 in the question. $\endgroup$ – Philipp Imhof Mar 21 '18 at 9:42
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    $\begingroup$ I would rather say: "if we combine $D$ and $F$" instead of "...$P(D)$ and $P(F)$". $\endgroup$ – drhab Mar 21 '18 at 10:00
  • $\begingroup$ Good point, thanks. $\endgroup$ – Philipp Imhof Mar 21 '18 at 10:01

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