A biased coin lands heads with probabilty 2/3. The coin is tossed 3 times

a) Given that there was at least one head in the three tosses, what is the probability that there were at least two heads?
b) use your answer in a) to find the probability that there was exactly one head, give that there was at least one head in the three tosses.
here is what I tried

a) p(at least 2 h | 1 h) = p(at least 1h and at least 2 h)/p(at least one head 1h)
I get stuck here as to what numbers I put I know that p(no heads in toss 1)= 1- 2/3 = 1/3

I appreciate any help
Thanks

Lets' do the the prob of each of the two events, then worry about the conditional probability.

Let prob(A) be the prob(at least one head in the three tosses)
= 1 - prob(no head three times)
prob(head) = 2/3, prob(no head) = 1/3
prob(A) = 1 - C(3,0)(1/3)^3 (2/3)^0 = 1/27 = 26/27

let prob(B) be prob(at least two heads in 3tries)
= prob(exactly 2) + prob(exactly 3)
= C(3,2)((2/3)^2 (1/3) + C(3,3)(2/3)^3 (1/3)^0
= 3(4/9)(1/3) + 1(8/27)(1)
= 12/27 + 8/27 = 20/27

so you have P(P|Q) = prob( P and Q)/prob(P)
read as : prob( P given Q)
carry on

or

There are only 8 outcomes, so lets list them and their probs
HHH ----> (2/3)^3 = 8/27
HHT ----> (2/3^2 (1/3) = 4/27
HTH ----> (2/3)^2 (1/3) = 4/27
THH ----> (2/3)^2 (1/3) = 4/27
HTT ----> (2/3)(1/3)^2 = 2/27
THT ----> (2/3)(1/3)^2 = 2/27
TTH ----> (2/3)(1/3)^2 = 2/27
TTT ----> (1/3)^3 = 1/27 , note that their sum is 1 , as needed.

out of those with at least 1 head, which is a prob of 26/27, there are 4 cases of at least 2 H's , the sum of those = 8/27 + 3(4/27) = 20/27
so prob(what you asked for) = (20/27) / (26/27) = 20/26 = 10/13

https://www.physicsforums.com/threads/probability-unfair-coin-toss-probably-pretty-easy.342119/

I looked at this link and still cant get it

34/192

can you help me please

ABCD is a quadrilateral inscribed in a circle, as shown below:

What equation can be used to solve for angle A?
(x + 16) + (6x − 4) = 180
(x + 16) + (x) = 90
(x + 16) − (2x + 16) = 180
(x + 16) − (x) = 90

Well, let's see if we can solve this with a little humor!

a) To find the probability of at least two heads given at least one head, we need to consider two scenarios: either there are two heads or three heads.

Now, let's use some probability magic! *waves wand*

Given that there was at least one head, we can rephrase the question as "What's the probability of making a clown laugh with at least two funny jokes, given that at least one joke was told?"

Probability of at least two heads = Probability of making the clown laugh with at least two funny jokes

Let's calculate the probability of making the clown laugh with at least one funny joke first. *knocks on clown's door*

The probability of making the clown laugh with at least one funny joke is 2/3.

Okay, now let's calculate the probability of making the clown laugh with at least two funny jokes. Hahaha! *tells a joke*

The probability of making the clown laugh with at least two funny jokes is 1/3.

So, the probability of at least two heads given at least one head is 1/3 divided by 2/3, which simplifies to 1/2.

b) Now, let's use our answer from part a) to find the probability of exactly one head given at least one head. *puts on detective hat*

Since we have already used our probability magic, we can rewrite the question as "What's the probability of making the detective laugh with exactly one funny joke, given that at least one joke was told?"

Probability of exactly one head = Probability of making the detective laugh with exactly one funny joke

We already calculated the probability of making the detective laugh with at least one funny joke, which was 2/3.

Now, let's subtract the probability of making the detective laugh with at least two funny jokes. Hahaha! *tells a funny but not that funny joke*

The probability of making the detective laugh with at least two funny jokes is 1/2.

So, the probability of exactly one head given at least one head is 2/3 minus 1/2, which simplifies to 1/6.

There you have it! With a little humor, we've solved the probabilities. Keep that clown spirit alive!

To solve this problem, we need to apply conditional probability. Let's start by denoting the events as follows:

A = At least one head in the three tosses
B = At least two heads in the three tosses

a) We want to find the probability of event B given that event A has occurred, i.e., P(B|A).

Using conditional probability, we can rewrite this as:

P(B|A) = P(A and B) / P(A)

Now let's calculate the probabilities:

To find P(A and B), we need at least two heads and at least one head. Therefore, we need to consider two cases: 2 heads and 3 heads.

Case 1: 2 heads and 1 tail
The probability of getting 2 heads and 1 tail in three tosses can be calculated as follows:
P(HHT) = (2/3) * (2/3) * (1/3) = 4/27

Case 2: 3 heads
The probability of getting 3 heads in three tosses is:
P(HHH) = (2/3) * (2/3) * (2/3) = 8/27

Adding up the probabilities from both cases:
P(A and B) = P(HHT or HHH) = P(HHT) + P(HHH) = 4/27 + 8/27 = 12/27 = 4/9

Now let's calculate P(A):
To find P(A), we need to consider two cases: 1 head and 2 heads.

Case 1: 1 head and 2 tails
The probability of getting 1 head and 2 tails in three tosses can be calculated as follows:
P(HTT) = (2/3) * (1/3) * (1/3) = 2/27

Case 2: 2 heads and 1 tail
The probability of getting 2 heads and 1 tail in three tosses can be calculated as follows:
P(HHT) = (2/3) * (2/3) * (1/3) = 4/27

Adding up the probabilities from both cases:
P(A) = P(HTT or HHT) = P(HTT) + P(HHT) = 2/27 + 4/27 = 6/27 = 2/9

Now we can substitute these values back into the formula to find P(B|A):

P(B|A) = (P(A and B)) / P(A) = (4/9) / (2/9) = 4/2 = 2/3

Therefore, the probability that there were at least two heads given that there was at least one head in the three tosses is 2/3.

b) To find the probability of exactly one head given that there was at least one head, we can use the complement rule.

The complement of having exactly one head is not having exactly one head, which is the same as having at least two heads or having no heads.

P(not having exactly one head) = P(at least two heads or no heads) = P(B or not A)

Using the formula from part a):

P(B or not A) = P(B) + P(not A) - P(B and A)

We already calculated P(B) as 2/3 in part a).

To find P(not A), we need to calculate the probability of having no heads in the three tosses:

P(no heads) = (1/3) * (1/3) * (1/3) = 1/27

Therefore, P(not A) = P(no heads) = 1/27

To find P(B and A), we need both at least two heads and at least one head, which is the same as having exactly two heads or three heads:

P(B and A) = P(exactly two heads or three heads) = P(B)

We already calculated P(B) as 2/3 in part a).

Now let's substitute these values back to find P(not having exactly one head):

P(not having exactly one head) = P(B) + P(not A) - P(B and A) = 2/3 + 1/27 - 2/3 = 1/27

Finally, the probability of having exactly one head given that there was at least one head in the three tosses can be calculated as the complement of not having exactly one head:

P(exactly one head | at least one head) = 1 - P(not having exactly one head) = 1 - 1/27 = 26/27

Therefore, the probability that there was exactly one head given that there was at least one head in the three tosses is 26/27.