please help, i think im right need reassurance On the exam, without them, or I w

Question

please help, i think im right need reassurance

On the exam, without them, or I won't make you evaluate your expressions to get a final answer. calculators will be allowed, but I'll take care that all questions can be done not 1. Suppose that out of a group of 12 students, there are four from Staten Island, four from Brooklyn, and four from Manhattan. A group of four students are chosen at random. What is the probability that this group includes members from all of Staten Island, Brooklyn, and Manhattan? 2. In cases of ear infections in a certain country, it is treated with penicillin 20% of the time, with amoxicillin 60% of the time, and is left untreated the rest of the time. These treatments succeed treatment penicillin amoxicillin no treatment success rate 90% 95% 50% at the following rates: Suppose a patient's ear infection has not been cured. What's the probability that she received no treatment? 3. The number of days of rain over this weekend is modelled as a random process with sample space [0, 1, 2. Suppose that the probability of no days of rain is .5, the probability of one day of rain is .3, and the probability of two days of rain is .2. List every possible event given this sample space, and state the probability of each event. 4. On a randomly selected day in August, the Mets have a 1/4 chance of playing a game at home and a 3/4 chance of playing on the road. They win with probability 4. Conditional on playing at home, they win with probability .5. Conditional on playing on the road, how likely are they to win? 5. Sarah has a biased coin that is heads with probability 1/3 and tails with probability 2/3. She makes three independent coin flips in sequence (a) What is the probability that all three flips are heads? (b) 3 What's the probability of obtaining at least one result of heads?

Explanation / Answer

Out of the group of 12 students, 4 are from Staten Islands, 4 are from Brooklyn and 4 from Manhattan

The number of ways of creating a single group of 4 from these 12 is equal to the binomial coefficient C(12,4)=12!/4!(12-4)!

That is, C(12,4)=12!/4!8!=495

If we want a group that includes at least one member from Staten Island, Brooklyn, and Manhattan, we first select one person from State Island from the 4 people in 4 ways. Next, we select the person from Brooklyn in another 4 ways and finally, the person from Manhattan can be selected in another 4 ways. We are free to choose another member from the remaining 12-3=9 people without restriction. Thus, this member can be chosen in 9 ways

The total number of ways of selecting our group is 4*4*4*9=576 but this takes into account the order of people

The number of ways of selecting our group is therefore 576/4! = 576/24=24

The required probability is therefore 24/495 = 0.04848....

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