Fair vs. biased coin probability

[magnifik]

Homework Statement

A 3-bit character consists of 0's and 1's. The values of the bits are determined by tosses of a fair coin.
(a) Find the probability that the first bit is 1.
(b) Find the probability that the first bit AND the third bit are 1.
(b) Find the probability that the first bit is 1 if the coin is biased.

The Attempt at a Solution

(a) There are 8 possible combinations of bits. Of these 8, 4 of them have 1s as the first digit so I get P(A) = 4/8. I am confused though.. is this the right logic? I thought about it another way where P(A) = 1/2 because the coin can land on heads or tails, so there is an equal chance that it would be either 0 or 1. I know both methods of thinking give the same answer, but I'm not sure which is most correct.

(b) Again, since I'm unsure about the method I'm not sure what the correct way to go about this is. I get P(B) = 2/8 from looking at this set {000, 001, 010, 011, 100, 101, 110, 111}. I get P(B) = (1/2)*(1/2) = 1/4 if I do it based on coin flip (I don't think it's mutually exclusive because they could both be 0 or both be 1, but I might be incorrect). Again both are equal.

(c) I don't know where to start on this one because the question doesn't state any values for the bias. Is there a general rule for this?

Any help would be appreciated! Thanks.

 

[cepheid]

For part (a), the both methods are correct. Your second method is correct because each coin toss is 50-50, and its outcome is completely independent from the other coin tosses. That having been said, your first method is also correct it must be true that 50% of the total possible outcomes of 3 consecutive coin tosses must have 1 as the outcome of the first one, in order to be consistent with the above. The way to think about it is as follows: there are 2 possible outcomes for the first coin toss, and then for each of those two possibilities, there are two possible outcomes for the second coin toss. As a result, there are 2*2 = 2² possible outcomes of two consecutive coin tosses. Two of these four outcomes have a 1 as the result of the first toss, and two of them have a 0.

Similarly, for each of the possible outcomes of the two coin tosses, there are two possible outcomes for the third one. As a result, are 2*2*2 = 2³ possible outcomes. Four of these eight started with a 1 and the other four started with a 0. So you can see how everything is consistent.

For (b), since all 8 outcomes are equally likely, just count up the number of outcomes in which the the first AND third bits are 1, and divide this by the total number of outcomes. EDIT: but your second method is fine as well. Since your coin tosses are independent events, you can multiply their probabilities so that

P(1st toss = 1 AND 3rd toss = 1)

= P(1st toss = 1)*P(2nd toss = 1 OR 0, don't care which)*P(3rd toss = 1)

= (1/2)*(1)*(1/2) = 1/4

For (c), the principle is the same, except the tosses are no longer 50-50.

 

[magnifik]

Ok, thanks. Say for the case of the union of each bit being a 1. Is the following correct...

There are 2^3 = 8 possible outcomes. Seven out of eight of these have a 1 in either the first, second, or third position (or any combination of the three as that is how I understand a union to be). So is it right to say that the probability is 7/8?

 

[cepheid]

Ok, thanks. Say for the case of the union of each bit being a 1. Is the following correct...

There are 2^3 = 8 possible outcomes. Seven out of eight of these have a 1 in either the first, second, or third position (or any combination of the three as that is how I understand a union to be). So is it right to say that the probability is 7/8?

Yeah, if you want the probability that the first bit is 1 OR the second bit is 1 OR the third bit is 1 and it is NOT an exclusive OR (i.e. any two could be 1, or all three of them), then again just counting the number of desired outcomes and dividing by the number of possible outcomes is a completely valid method. In fact, this is always a valid method IF every outcome is equally likely.

Another simple way to do this would be to compute the probability of the complement (opposite) outcome and subtract it from 1 so that:

P(1st =1 OR 2nd = 1 OR 3rd = 1) = 1 - P(none are 1) = 1 - 1/8 = 7/8.

By the way, I edited my previous post, not sure if you caught it.

 

[magnifik]

What if value of bias is unknown (as in this case)?
I'm not sure what values to use instead of 1/2

 

[cepheid]

What if value of bias is unknown (as in this case)?
I'm not sure what values to use instead of 1/2

If they didn't give you values for the probability of flipping a 1 or a 0, they must just want an algebraic answer. Call the probability of a 1, 'p', and the probability of a 0, '1-p' or something like that.