We have two coins, A and B. For each toss of coin A, we obtain Heads with probability 1/2; for each toss of coin B, we obtain Heads with probability 1/3. All tosses of the same coin are independent. We select a coin at random, where the probabilty of selecting coin A is 1/4, and then toss it until Heads is obtained for the first time.
The expected number of tosses until the first Heads is:
11/4
Ah, the world of probability, where heads and tails reign supreme! Now, let's calculate the expected number of tosses until the first heads pops up.
Let's break it down. We have a 1/4 chance of selecting coin A and a 3/4 chance of selecting coin B.
If we select coin A, the probability of obtaining heads on the first toss is 1/2. So, on average, we'll need 2 tosses (1 toss for tails and then 1 toss for heads).
If we select coin B, the probability of obtaining heads on the first toss is 1/3. This means we'll need an average of 3 tosses (1 toss for tails and then 2 tosses for heads).
Now, let's bring it all together using some fancy math skills! The expected number of tosses is:
(1/4 * 2) + (3/4 * 3) = 1/2 + 9/4 = 11/4
So, the expected number of tosses until the first heads is 11/4 or 2.75 tosses. And remember, sometimes in life, you gotta toss a few coins to get ahead!
To calculate the expected number of tosses until the first Heads is obtained, we can use the concept of conditional probability.
Let's consider the expected number of tosses E, starting with coin A and then switch to coin B when a Tails is obtained.
When starting with coin A, the probability of obtaining a Heads on the first toss is 1/2, and the expected number of tosses after that would be E. So the expected number of tosses when starting with coin A is 1 + E.
When starting with coin B, the probability of obtaining a Heads on the first toss is 1/3, and the expected number of tosses after that would also be E. So the expected number of tosses when starting with coin B is 1 + E.
Given that the probability of selecting coin A is 1/4 and coin B is 3/4, we can use these probabilities to calculate the overall expected number of tosses.
E = (1/4)*(1 + E) + (3/4)*(1 + E)
Simplifying the equation:
E = (1/4) + (3/4)E + 1/4 + (3/4)E
E = 1 + (3/2)E
To solve for E, we can subtract (3/2)E from both sides and subtract 1 from both sides:
E - (3/2)E = -1
-1/2 E = -1
Finally, divide both sides by (-1/2) to solve for E:
E = -1 / (-1/2)
E = -1 * (-2/1)
E = 2
So, the expected number of tosses until the first Heads is obtained is 2.
To find the expected number of tosses until the first Heads is obtained, we can use the concept of expected value.
Let's define the random variable X to represent the number of tosses until the first Heads is obtained.
There are two possible scenarios:
1. We select coin A: In this case, the probability of obtaining Heads on the first toss is 1/2. If we don't get Heads on the first toss, we start again and the expected number of additional tosses until we get Heads is still X. So for this scenario, the expected number of tosses until the first Heads is obtained is 1 + X/2.
2. We select coin B: In this case, the probability of obtaining Heads on the first toss is 1/3. If we don't get Heads on the first toss, we start again and the expected number of additional tosses until we get Heads is still X. So for this scenario, the expected number of tosses until the first Heads is obtained is 1 + X/3.
Now, we need to consider the probability of selecting each coin. We are told that the probability of selecting coin A is 1/4 and the probability of selecting coin B is 3/4.
Therefore, we can calculate the expected number of tosses as follows:
Expected number of tosses = (1/4) * (1 + X/2) + (3/4) * (1 + X/3)
= 1/4 + X/8 + 3/4 + X/4
= 1 + (X/8) + (3/4) + (X/4)
= 1 + (X/8) + (3 + 2X)/4
To simplify this expression, we can put the terms with X over a common denominator:
Expected number of tosses = 1 + (4X + 8X + 6) / 32
= 1 + (12X + 6) / 32
= 1 + (3X + 1) / 8
So the expected number of tosses until the first Heads is obtained is 1 + (3X + 1) / 8.