Calculating Probabilities of Coin Tosses

[kent davidge]

Homework Statement

A coin is thrown 8 times. What's the probability for obtaining

a) 5 heads

b) at least 1 head

c) at most 2 heads

Homework Equations

Bayer's rule, I think.

The Attempt at a Solution

Using Bayer's rule:

a) $P(5 \text{he}, 3 \text{ta}) = P(5 \text{he} | 3 \text{ta}) P(3 \text{ta}) = (1/2)^5 (1/2)^3 = (1/2)^8$;

b) $\sum_{i=1}^{8} P(i \text{he}, (8-i) \text{ta}) = \sum_{i=1}^{8} P(i \text{he} | (8-i) \text{ta}) P((8-i) \text{ta}) = \sum_{i=1}^{8} (1/2)^i (1/2)^{8-i} = 8(1/2)^8$.

c) $\sum_{i=0}^{2} P(i \text{he}, (8 - i) \text{ta}) = 3 (1/2)^8$.

Is this correct?

 

[member 587159]

(a) Is wrong
(b) Is wrong.

However, it is much easier to use the complement:

P(at least 1 head) = 1-P(no heads) = 1 - (1/2)^8

(c) Wrong too.

P(at most 2 heads) = P(0 heads) + P(1 heads) + P(2 heads)

and the probability on P(1 heads) or P(2 heads) is NOT equal to (1/2)^8, as you seem to think.

Think about it intuitively: according to your answer, the probability that there is at least 1 head is (1/2)^8 = 3%. You can feel that the probability must be a lot higher, near 100%.

In how many ways can we have 1 heads if we throw 8 times?
In how many ways can we have 2 heads if we throw 8 times?

You should take this into account! This is the same reason that (a) - (b) were wrong too.

 

[tnich]

Homework Statement

A coin is thrown 8 times. What's the probability for obtaining

a) 5 heads

b) at least 1 head

c) at most 2 heads

Homework Equations

Bayer's rule, I think.

The Attempt at a Solution

Using Bayer's rule:

a) $P(5 \text{he}, 3 \text{ta}) = P(5 \text{he} | 3 \text{ta}) P(3 \text{ta}) = (1/2)^5 (1/2)^3 = (1/2)^8$;

b) $\sum_{i=1}^{8} P(i \text{he}, (8-i) \text{ta}) = \sum_{i=1}^{8} P(i \text{he} | (8-i) \text{ta}) P((8-i) \text{ta}) = \sum_{i=1}^{8} (1/2)^i (1/2)^{8-i} = 8(1/2)^8$.

c) $\sum_{i=0}^{2} P(i \text{he}, (8 - i) \text{ta}) = 3 (1/2)^8$.

Is this correct?

I think you mean Bayes' rule, but that is not appropriate in this case. What you need to use here is the binomial distribution.

 

[member 587159]

I think you mean Bayes' rule, but that is not appropriate in this case. What you need to use here is the binomial distribution.

EDITED:

I agree that this problem perfectly lends itself for an application for binomial distribution, but maybe it is interesting for the OP to do it the hard way.

 

[tnich]

Bayes' rule here is appropriate, in the sense that it is the same as the product rule for independent events. So the OP isn't actually doing anything wrong here.

I agree that the events are independent and that $P(5he | 3ta) = P(5he)$.
But I think of Bayes rule as
$P(B|A) = \frac {P(A|B) P(B)} {P(A)}$
which the OP has not applied in his post, nor can I think of a useful way to do so.
In addition, the OP has not arrived at the correct answer for part a) or any of the others, so his attempt to use conditional probabilities has led him down the wrong path.
If the question had been, "what is the probability that the first 5 throws are heads and the rest tails", then his approach would have been correct.

 

[member 587159]

I agree that the events are independent and that $P(5he | 3ta) = P(5he)$.
But I think of Bayes rule as
$P(B|A) = \frac {P(A|B) P(B)} {P(A)}$
which the OP has not applied in his post, nor can I think of a useful way to do so.
In addition, the OP has not arrived at the correct answer for part a) or any of the others, so his attempt to use conditional probabilities has led him down the wrong path.
If the question had been, "what is the probability that the first 5 throws are heads and the rest tails", then his approach would have been correct.

Yes, the OP must take into account that it is possible to obtain 5 heads in more than 1 way. And I thought of Bayes' rule more like $P(A \cap B) = P(A)P(B|A)$, but you are right that this isn't even Bayes'rule (although this formula can be used to derive Bayes' rule), so you can ignore my comment on your post.

 

[kent davidge]

I thought Bayes' rule could be used in any situation involving probability. What should I use here

 

[tnich]

I thought Bayes' rule could be used in any situation involving probability. What should I use here

The binomial distribution

 

[PeroK]

I thought Bayes' rule could be used in any situation involving probability. What should I use here

You are fundamentally misunderstanding Bayes' theorem in this case.

If you toss a coin $8$ times, then exactly $5$ heads and exactly $3$ tails are two descriptions of the same outcome. So:

$P(5 heads | 3 tails) = 1$

 

[tnich]

The binomial distribution

I suggest that you do some research on Bayes' theorem (Bayes' rule). That should give you a better idea of how and when to apply it.
What you have been trying to do is use conditional probabilities. When the events you are considering are independent, as in the case of tosses of a fair coin, the conditional probability of tossing H given previous tosses TTT is P(H|TTT) = P(H) = $\frac 1 2$. The outcomes of previous tosses have no effect on the outcome of the next toss, so conditional probabilities do not help you in this case.
What you need to consider is how many ways there are to toss 5 heads and 3 tails. You could toss HHHHHTTT or HHHHTHTT or HHHHTTHT or . . . Each of the these ways would have probability $(\frac 1 2)^8$.

 

[kent davidge]

...exactly $3$ tails are two descriptions of the same outcome. So: $P(5 heads | 3 tails) = 1$

I think your input reads "the probability for obtaining 5 heads given that I obtained 3 tails before". If that's it, I don't understand how it can be equal to 1. There's no 100% certainty that I will get 5 heads in 5 tosses.

The binomial distribution

I will have a look at the use of binomial distribution in probabilities.

I suggest that you do some research on Bayes' theorem (Bayes' rule). That should give you a better idea of how and when to apply it.
What you have been trying to do is use conditional probabilities. When the events you are considering are independent, as in the case of tosses of a fair coin, the conditional probability of tossing H given previous tosses TTT is P(H|TTT) = P(H) = $\frac 1 2$. The outcomes of previous tosses have no effect on the outcome of the next toss, so conditional probabilities do not help you in this case.
What you need to consider is how many ways there are to toss 5 heads and 3 tails. You could toss HHHHHTTT or HHHHTHTT or HHHHTTHT or . . . Each of the these ways would have probability $(\frac 1 2)^8$.

I see, I think. So in particular there are 8! possibilities in this case.

EDIT: Ops, 8! / (5! 3!), actually.

 

[kent davidge]

Oh I realized how the binomial theorem comes into our case. For exactly five heads we have: $(8! / (5! 3!)) (1/2)^8 \approx 21 \%$; At most two heads: $(8! / 8! + 8! / (1! 7!) + 8! / (2! 6!))(1/2)^8 \approx 14 \%$...

 

[tnich]

Oh I realized how the binomial theorem comes into our case. For exactly five heads we have: $(8! / (5! 3!)) (1/2)^8 \approx 21 \%$; At most two heads: $(8! / 8! + 8! / (1! 7!) + 8! / (2! 6!))(1/2)^8 \approx 14 \%$...

Yes. Now how do you the probability of at least one heads?

 

[FactChecker]

I thought Bayes' rule could be used in any situation involving probability. What should I use here

You certainly can use Bayes' rule as part of the solution here but it probably doesn't help you. Bayes' rule is very useful when you are given information and want to see how it changes probabilities. So you could imagine that someone told you there were 3 tails and then you use Bayes' rule to figure out P(5he|3ta), as you indicated in Post #1a. But that may just complicate things in this problem. Use Bayes' rule when you are specifically asked for the conditional probability of 5ha, given 3ta, or when it really simplifies the solution because P(5he|3ta) is obvious for some reason.

 

[Ray Vickson]

I think your input reads "the probability for obtaining 5 heads given that I obtained 3 tails before". If that's it, I don't understand how it can be equal to 1. There's no 100% certainty that I will get 5 heads in 5 tosses.

I will have a look at the use of binomial distribution in probabilities.

I see, I think. So in particular there are 8! possibilities in this case.

EDIT: Ops, 8! / (5! 3!), actually.

You could use Bayesian ideas in this problem (known as a conditioning argument) but it would not help much. For example, in 8 tosses, we have
$$P_8(5\; \text{heads})= P_8(5\; \text{heads} | \text{head first} ) P(\text{head first})
+ P_8(5\; \text{heads} | \text{tail first} ) P(\text{tail first}) $$
Now, given that a "head" is first we have left 7 tosses and need 4 heads; given that a "tail" is first, we have left 7 tosses and need 5 heads. Since the remaining tosses do not depend on the first toss, we have
$$P_8(5\; \text{heads} | \text{head first} ) = P_7( 4 \; \text{heads})$$
and
$$P_8(5\; \text{heads} | \text{tail first} ) = P_7(8 \; \text{heads})$$
So, we replace the original 8-toss problem by two similar 7-toss problems. This is really not terribly helpful!

 

[PeroK]

I think your input reads "the probability for obtaining 5 heads given that I obtained 3 tails before". If that's it, I don't understand how it can be equal to 1. There's no 100% certainty that I will get 5 heads in 5 tosses.

The notation $P(A|B)$ means the probability of event $A$ given that we have event $B$. It is not about a sequence of things.

For example. If you roll a die and event $A$ is that the die reads $6$ and event $B$ is that the die reads an even number, then

$P(A|B) = P(6|even) = 1/3$

Which means that if you know it's an even number, then there is a probability of $1/3$ that it is a six.

It doesn't mean that if you roll a die once and it is even and then you roll it again, the chance of a 6 on the second roll is $1/3$.

In your case, if you toss a coin 8 times and event $A$ is exactly 5 heads and event $B$ is exactly 3 tails, then

$P(A|B) = 1$

Given you have exactly 3 tails, you must have exactly 5 heads.

Although, in this case you could say that $A = B$ as they are the same event.

 

[kent davidge]

Thanks to all for your replies. To check if I understood your latest replies, I considered the following. Consider a box with two green and two white balls. What's the probability for picking, $G,G,B,B$ in this order? And for picking $G,B,G,B$? Using your hints the probabilities are $1/6$ for either case. That has to be so because if I asked what's the probability for picking $G,G,B,B$ in any order, the answer must be $1$, because then I'd be asking what's the probability for picking all the balls from the box, and there are exactly $4! / (2! 2!)$ ways of doing so, which cancels the $6$ in the denominator.

It's remarkable how counter-intuitive it is that the probabilities are all equal in this case. I would say at first that it's more difficult to have $G,G,B,B$ in this order than $G,B,G,B$. The latter seems more random!