Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p , where 0<p<1.
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.
We are interested in calculating the probability that there are 5 Heads in the first 6 tosses and 3 Heads in the last 5 tosses. Give the exact numerical values of a,b,c,d that would match the answer ap7(1-p)3+bpc(1-p)d.
Thanks !!
To determine the probability that the second head occurred at the 4th toss, given that there were 4 heads in the first 7 tosses, we need to consider the number of ways this outcome can occur.
First, let's break down the possibilities for the first 7 tosses:
HHHHHTT
HHHHHHT
HHHHHTH
HHHHH
...
...
(There are many more possibilities)
Out of all these possibilities, we need to focus on those where the second head occurs at the 4th toss. Looking at the above possibilities, we find that this happens in the first three cases: HHHHHTT, HHHHHHT, HHHHHTH.
Thus, there are 3 possibilities where the second head occurs at the 4th toss.
Now, let's find the total number of possibilities for the first 7 tosses. Since each toss can have two outcomes (heads or tails) and there are 7 independent tosses, there are 2^7 = 128 possible outcomes.
Therefore, the probability that the second head occurred at the 4th toss, given that there were 4 heads in the first 7 tosses, is 3/128.
Now, let's move on to the second question regarding the numerical values of a, b, c, and d.
We are interested in calculating the probability that there are 5 heads in the first 6 tosses and 3 heads in the last 5 tosses. To do this, we can use the binomial probability formula.
The binomial probability formula for getting k successes in n independent trials, where the probability of success in each trial is p, is:
P(k out of n) = (n choose k) * (p^k) * ((1-p)^(n-k))
In this case, k = 5, n = 6, and p = probability of getting a head on each toss. So we have:
P(5 out of 6) = (6 choose 5) * (p^5) * ((1-p)^(6-5))
Simplifying this expression, we find:
P(5 out of 6) = 6p^5(1-p)
Similarly, for the second part of the probability, where we want 3 heads in the last 5 tosses, we have:
P(3 out of 5) = (5 choose 3) * (p^3) * ((1-p)^(5-3))
Simplifying, we have:
P(3 out of 5) = 10p^3(1-p)^2
Putting it all together, the overall probability will be given by:
ap^7(1-p)^3 + b
where a = 6, b = 10, c = 5, and d = 2.
So, the numerical answer would be:
6p^7(1-p)^3 + 10p^3(1-p)^2