Halp with conditional probability questions and what not Watch

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So I'm trying to study for an exam (equivalent to A level) which has probability and we were given a few pages to practice on answers included. But I'm just retarded like that especially when it comes to probability. :/ could someone pleasee help me out with how to get these answers..?

1. Susan takes exams in Maths, French and History. The probabilities that she passes are 0.7, 0.8 and o.6 respectively. Given that her performances in each subject are independent, find the probability that Susan
a) fails all examination (0.024) -that one i understood
b) fails just one examination (0.452) i got this one in the end but is there a proper method..?
c) given that susan fails just one examination, find the probability that it is History that she fails (0.4956) wat. plz help.

2. Four coins are biased so that probabilites of getting head are 0.4, 0.48, 0.52 and 0.6. One coin is chosen at random, and, on being tossed three times gives three heads. What is the probability that the coin chosen was the coin with the bias 0.52? (2197/8300)

3. On a certain Pacific Island, a particular disease is caught by one person in a thousand. 98% of those who have the disease respond positively to the diagnostic test, but 3% of those who do not have the disease also respond positively to the same test. If a person is selected at random and responds positively, what is the probability he actually has the disease? (0.03166)

4. Two dices are rolled. Given that there is at least one six, what is the probability that both are sixes? (1/11) pathetic I know.

5. If there are 23 people in a room, find the probability that at least two share the same birthday. (0.50729...)

6. Research in a Latvian shows that the probability that a boy is left handed is 0.14 and the probability that a girl is left handed is 0.08. In a Latvian school of 1000 students there are 650 girls and 350 boys.
this one's particularly important and there's no answer ^^ --->Two students are selected at random. Find the probability that at least one is left handed.

Then there's some stuff on permuations, combinations and sequences

7. A class of 9 students contains 4 girls and 5 boys (two of which are twin brothers). Three students are to be chosen to represent the class in a maths competition.
a. In how many different ways can the three students be selected from the class (9C3)
After reading the rules of the competition the teacher realizes that each team must contain at least one boy and at least one girl.
b. In how many ways can the three students be selcted if there must be at least one boy and at least one girl?
c. The teacher chooses a team at random from all those teams that contain at least one boy and at least one girl. Find the probability that the selected team contains both of the twin brothers.

8. A ball is dropped from a horizontal plane and rebounds succesively. The height above the plane reached by the ball after the first impact is 4m. After each impact the ball rises to 3/4 of the height reached after the previous impact. Calculate the total vertical distance travelled from the first impact until the ball comes to rest.

Pleaseee help me ill love you forever in a totally not creepy way ^_^
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Report 4 years ago
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Report 4 years ago
1b) use Theorem of Total Probability. P(fails one examination) = sum of P(fails one examination given that it is X) where X is either French, History or Maths so three terms to find and add
1c) use Bayes' Theorem P(A|B) = P(A)P(B|A) / P(B) where A = "fails history" and B = "fails one examination". You should get P(B|A) in your working in 1b)

more to come...

2) "Random" means the four coins have equal chance of being selected. So find P("3 heads") for each coin (= p^3 where p = "probability of one head"), add them together. Your answer is equal to P("3 heads"|"0.52 biased coin") / sum of P("3 heads")

3) Again, Bayes' Theorem. So A = "has the disease", B = "test is positive" You are given P(B|A) = 0.98 and P(B|A') = 0.03. Your task is to find P(A|B) (hint: find P(B) using Theorem of Total Probability)

4) P(X = 6 ∩ Y = 6 | X = 6 ∪ Y = 6). You know P(X = 6 ∩ Y = 6) is 1/36. You can get P(X = 6 ∪ Y = 6) by inclusion-exclusion, the rest is up to you...
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Report 4 years ago
(Original post by AllyMew)
So I'm trying to study for an exam

There is an awful lot of maths there (including some mechanics in the last question)

Are you getting any support for this exam or are you self studying?

What is your previous level of maths and how long is it since you did any maths?
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Report 4 years ago
Most of these can be done using a small or large tree diagram. Many probability problems of this type I solve by tree diagram, even if I don't draw it out (or don't draw out every branch).

Bayes's Theorem: I like Yudkowsky's intuitive explanation, although the degree to which you find it "excruciatingly gentle" (as he puts it) varies wildly per person.

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