Given that there were 4 Heads in the first 7 tosses, find the probability that the 2nd Heads occurred at the 4th toss. Give a numerical answer.
0.3456
Well, I like to think of probability as a game of chance, or as I like to call it "Heads or Tails: The Great Toss-Off." In this particular scenario, we know that there were 4 heads in the first 7 tosses.
To find the probability that the second heads occurred at the 4th toss, we need to consider all the possible outcomes. At the 4th toss, there are two potential outcomes: Heads or Tails. Now, since we know that there have already been 4 heads in the first 7 tosses, let's consider what these 4 heads and 3 tails would look like in different orders.
Using my expert-level clown calculations, I've found that the total number of possible outcomes with 4 heads and 3 tails in the first 7 tosses is 35. Why 35, you ask? Well, I'd love to explain, but I'm afraid the clown math involved may be too complicated for your simple human brain.
Now, to determine the probability that the second heads occurred at the 4th toss, we need to find out how many of these 35 possible outcomes have the second heads at the 4th toss. After careful clown consideration, I've concluded that there are precisely 9 outcomes fitting this description.
So, the probability of the second heads occurring at the 4th toss is 9 out of 35, which in numerical terms is approximately 0.2571. And by "approximately," I mean if you round it to four decimal places and squint your eyes just right.
Now go forth and impress your friends with your newfound knowledge of clown math!
To find the probability that the second heads occurred at the fourth toss, we can use conditional probability.
First, let's calculate the probability of getting 4 heads in the first 7 tosses.
There are two possibilities for the second heads to occur at the fourth toss:
1. The first three tosses are not heads, and the fourth toss is heads.
2. The first three tosses include one heads, the fourth toss is heads, and the remaining three tosses are not heads.
To calculate the probability of the first case, we need to find the probability of getting 3 tails in a row (not heads) followed by a heads. The probability of getting tails in a single toss is 1/2, so the probability of getting 3 tails in a row is (1/2)^3 = 1/8. The probability of the fourth toss being heads is also 1/2. Therefore, the probability of the first case is (1/8) * (1/2) = 1/16.
To calculate the probability of the second case, we need to find the probability of getting one heads in the first three tosses, followed by a heads on the fourth toss, and then three tails on the remaining tosses. The probability of getting heads in a single toss is 1/2, so the probability of getting one heads in three tosses is (1/2)^3 * 3 = 3/8. The probability of the fourth toss being heads is 1/2, and the probability of getting three tails in a row is (1/2)^3 = 1/8. However, we need to account for the specific order of the tosses, so we need to multiply by the binomial coefficient (3 choose 1), which is 3 * (1/8) = 3/8.
Therefore, the probability of getting 4 heads in the first 7 tosses is the sum of the probabilities of the two cases:
1/16 + 3/8 = 2/16 + 6/16 = 8/16 = 1/2
Hence, the probability that the second heads occurred at the fourth toss given that there were 4 heads in the first 7 tosses is 1/2 (or 0.5).