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Problem 1. Tosses of a biased coin 7 points possible (graded) Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0 < p < 1. 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. (You can click on the "STANDARD NOTATION" button below.) 2. Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation. Remember not to use ! or combinations in your answer. 3. 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. 4. 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 ap^7(1 - p)^3 + bp^c(1 - p)^d. a = b = c = d =

          Problem 1. Tosses of a biased coin
7 points possible (graded)
Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0 < p < 1.
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. (You can click on the "STANDARD NOTATION" button below.)
2. Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation. Remember not to use ! or combinations in your answer.
3. 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.
4. 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 ap^7(1 - p)^3 + bp^c(1 - p)^d.
a =
b =
c =
d =

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Problem 1. Tosses of a biased coin
7 points possible (graded)
Consider 10 independent tosses of a biased coin with the probability of Heads at each toss equal to p, where 0 < p < 1.
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. (You can click on the "STANDARD NOTATION" button below.)
2. Find the probability that there are 3 Heads in the first 4 tosses and 2 Heads in the last 3 tosses. Express your answer in terms of p using standard notation. Remember not to use ! or combinations in your answer.
3. 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.
4. 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 ap^7(1 - p)^3 + bp^c(1 - p)^d.
a =
b =
c =
d =

Added by J V.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Keondre Parker

06/15/2023

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Step 1: Calculate the probability that the ninth toss results in heads given that there are exactly six heads in the first eight tosses. Show more…

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Transcript

00:01 All right, so the probability of b given a is going to be a probability of b and a over the probability of a.

00:13 So the probability of a is going to be 8 choose 6 and that's going to be times p to the power 6 times 1 minus p to the power 2.

00:34 And then for a and b, we also know, well, we know that this is going to be p because b is the event of the ninth toss results in heads.

00:49 So that's just going to give us p and then that's going to be times 8 choose 6 times probability of 6 times 1 minus p squared.

01:04 Because that's the first eight tosses.

01:15 Okay, so in this case we can simplify this a little bit, cancel out terms.

01:20 So in this case, this would just be p.

01:27 So therefore, the probability that the ninth toss heads results in heads, given that there are exactly six tosses in the first eight tosses, is just p.

01:38 So for 2, we want three heads in the first four tosses and two heads in the last three.

01:45 So press our answer in terms of p.

01:48 So probability of three heads and four tosses is just going to be 4 choose 3 times p cubed 1 minus p.

02:10 And then two heads and three tosses is going to be the same.

02:15 Or i'm sorry, yeah, two heads and last three tosses.

02:21 So two heads and three tosses, 3 choose 2, p squared 1 minus p.

02:32 So since these events are independent, we're going to have basically probability of event 1 and event 2.

02:46 So we'll just be multiplying the two events.

02:54 So we're going to get 4 choose 3 times p.

03:00 Well, 4 choose 3 times 3 choose 2 times this would be p to the fifth and times 1 minus p squared.

03:19 For 3, we want four heads in the first seven tosses.

03:24 Also given that there's four heads in the first seven tosses, find the probability that the second heads occurs on the fourth toss.

03:37 So in that case, that's going to be the probability of e given f.

03:44 So we can write that as the probability of f given e times the probability of e and then divide that by the probability of f...

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