Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0<p<1.
Let A be the event that there are 6 Heads in the first 8 tosses. Let B be the event that the 9th toss results in Heads.
Find P(B∣A) and express it in terms of p using standard notation
mit2020
4a 30 =a
4b 4=b
4c 8 =c
4d 2=d
1 = p
2 = 12*p^5*(1-p)^2
3 = 0.2571428571
4a = 15
4b = 60
4c = 6
4d = 4
4e = 10
4f = 5
Well, well, we've got ourselves a probability problem! Let's see what kind of fun we can have here.
So, we need to find the probability of event B (the 9th toss resulting in Heads) given that event A (having 6 Heads in the first 8 tosses) has occurred. To do that, we'll need to use conditional probability.
The probability of event B given event A is denoted as P(B|A), and it's calculated by dividing the probability of both events happening (A and B) by the probability of the event A on its own. In other words:
P(B|A) = P(A and B) / P(A)
Now, let's break it down.
The probability of event A is the probability of getting 6 Heads in 8 tosses. There are 8 choose 6 ways to arrange 6 Heads in 8 tosses, which gives us (8 choose 6) * p^6 * (1-p)^2.
The probability of event B is simply p since the 9th toss is independent of the first 8 tosses.
Now we need to find the probability of both A and B happening. In our case, that means getting 6 Heads in the first 8 tosses AND getting a Head in the 9th toss. Since these events are independent, we can multiply their probabilities:
P(A and B) = P(A) * P(B)
Plugging it all in:
P(B|A) = P(A and B) / P(A)
= (P(A) * P(B)) / P(A)
= (8 choose 6) * p^6 * (1-p)^2 * p / (8 choose 6) * p^6 * (1-p)^2
= p
Woohoo, there it is! The probability of event B given event A is equal to p. So, P(B|A) = p.
Hope this got your probability taste buds tingling!
To find the conditional probability P(B|A), we need to calculate the probability of event B occurring given that event A has already occurred.
First, let's find the probability of event A occurring. Event A is defined as having 6 heads in the first 8 tosses.
The probability of getting 6 heads in 8 tosses can be calculated using the binomial probability formula:
P(A) = C(n, k) * p^k * (1-p)^(n-k)
where n is the number of trials, k is the number of successful outcomes, and p is the probability of success.
In this case, we have n=8, k=6, and p as the probability of getting heads in a single toss.
Therefore, P(A) = C(8, 6) * p^6 * (1-p)^(8-6)
= 28 * p^6 * (1-p)^2
Next, let's find the joint probability of events A and B occurring. The joint probability P(A∩B) is the probability of 6 heads in the first 8 tosses AND the 9th toss resulting in heads.
Since the 9th toss is independent of the first 8 tosses, the probability of getting heads on the 9th toss is simply p.
Therefore, P(A∩B) = P(A) * P(B) = P(A) * p
Finally, let's calculate P(B|A). This can be found using the formula:
P(B|A) = P(A∩B) / P(A)
Plugging in the values, we get:
P(B|A) = P(A∩B) / P(A)
= (P(A) * p) / P(A)
= p
So, P(B|A) = p, where p is the probability of heads in each toss.