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(b) (6 marks) Peter flips fair coin n + / times independently Mary flips fair coin times independently: What is the probability that Peter gets more heads than Mary...

Question

(b) (6 marks) Peter flips fair coin n + / times independently Mary flips fair coin times independently: What is the probability that Peter gets more heads than Mary does? Here is & positive integer. (e) (8 marks) A is # Binomial random variable with paramelers and p. Find

(b) (6 marks) Peter flips fair coin n + / times independently Mary flips fair coin times independently: What is the probability that Peter gets more heads than Mary does? Here is & positive integer. (e) (8 marks) A is # Binomial random variable with paramelers and p. Find



Answers

Probability A coin is tossed ten times. The probability of heads on each toss is $0.5 .$ Evaluate each probability.
a. exactly 5 heads $\quad$ b. exactly 6 heads $\quad$ c. exactly 7 heads

All right, if a single coin is toast and what is the probability off getting a heads be fewer than two tails. Ah, see exactly three tails. Okay, so the simple Spence off tossing coin is head 02 Those are the only true possibilities that you can't have it now. What is the probability off getting its? It's one off, too. And the second question say's. What is the probability of getting fewer than to tells fewer than two tails less than to receive is it was what was you certainly get list on two teeth. Um, tent. Exactly three towels ability off getting t if you just said queen ones with their now high bonuses.

Since three corns are flip together toe The sample space contained eight elements. Had the head help had the pill had had the tail tail till till head still had tail? Still had had had had the deal till till still So there are eight cases in the sample space. Now we need to find operability that there is exactly one head. Now the cases when there is one head because physical toe had tail, tail, tail, tail, head and tail had still so there are three cases when there is exactly one had occurred. So operability off. Exactly one had is comes out. Toby three. Upon total number of cases, that is it top ability is comes out the tree upon it. Now we need toe. Find a probability that there is, at most two pills. So the guess is then, at most two tails occur Our head, head head. Had the deal had had a tail tail tail still had a tail had tail Dale had had had had because, though, because at most two tails means there will be one tail or to tail or zero tell. So there are seven guesses for at most two tails So rebel T is comes Probe lt four at the most two tails is comes out Toby seven Upon total number of cases, that is it that just prepared these comes out Toby seven upon it. Now we need to find a job. Lt Best there is at least one head So number of cases when there is at least one head that is means there is to head or one head or three heads. So the cases are head, head head Had the tail had the had the tail tail Still still had the still had the pill Still had had the had had a pill So there are seven cases for at least one head. So probability off getting at least one had is comes out Toby seven upon it. Now we need to find labrum lt forgetting exactly two tails. So the cases when exactly two tails are come are had tell tale tale Still had the tail had still so there are three cases when there is exactly two tails. So the preventative will be tree upon total number of cases that is it. So the probabilities comes out to be three upon it. So this is over

1 64 We have a bounce going and an unbalanced going. So our unbalanced going is 0.6. So are probabilities, or 0.6 point four to add upto one that would be heads that would be tales and then and or tales and bizarre fair coin hands hails their point. So here we have had tense Miriam had tails. This would be a 0.3 point six times 60.5 0.6 times point fine. This would be tales and 0.2 entails tales also point to So now answer the probability questions. Being both heads, heads head 0.3 being tens pales or kills head. So I'd be adding out. My 0.3 plus might wait too. I got to be 0.5 b. Neither is ahead. So we're looking at white too. Sales sale e has heads giving The 1st 1 is ahead when and head. So given the fact that a is But we have a 0.3 over 0.6, which is 0.5 Ah, and hand given be is head 0.6 and white too. That's all right. Hands head Given the no given he is ahead. This is 0.3. So hands heads is 0.3. Given that he is, the head is our denominator, which is 0.5. So that equals points six and then g, we're looking at probability. Yeah, on point A given exactly one. And so we ready given exactly one head is 10.3 plus point to which is point fine again that point.

In this problem. We have two coins, one is biased and one is unbiased. That's marked as ah B and B bar in my tree diagram. Ah, the unbiased coin obviously would have, ah, 50% chance of heads and tails, whereas the biased coin, which is the 1st 2 ranches, will have a 60% chance of heads and 40% chance of tales importante. It's asking us for the probability that it lands on heads. So we look at all the branches with heads in it. That will be 0.5 times 0.6 plus 0.5 times 0.5, and ah, put that my calculator that will give us 0.55 and part B. That's a conditional problem. So we want to find the probability that it's unbiased, given that it's landed on tails that will be probability of unbiased and tails over a probability of tales in the numerator that is the last branch little B, 0.5 times 0.5. Probability of tales would be all the branches that include tails, so that will be 0.5 times 0.4 plus 0.5 times 0.5. Uh, when we put this into a calculator, it will give us the probability of 0.55 ah six if we rounded to three significant figures.


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