Please help. I completed 1-6 but these 4 are hard; the wording confusing.
1) Jane is playing a game of Monopoly, if she throws six on her next turn she will land on community chest. If she throws eight she will land on Mayfair with a hotel (owned by another player). Of the 20 community chest cards I will send her to jail, I will fine her £100, will win her £20 and I will send her back to Old Kent Road. Calculate the probability that on the next turn
a) she lands on mayfair
b) she wins money
c) she loses money
d)she ends up in jail
e) she does not end up on mayfair or Old Kent road
2) A and b are independent events such that p(A)=2/3 and p(a n b)=1/3. Find
a) P(B)
b) P(A U B)
3) events A and B are such that p(A)=2/3, p(B)=7/12 and p(A n B)'=1/4. State whether A and B are
a)mutually inclusive
b)independent
4)two numbers are selected at random from the sets (2,4,6,8) and (1,3,5,7). Find the probability that
a) their sum is less than 8
b) the sum is even
c) the difference is 3
d) the product is even
ok, I will provide some hints, but you will have to work on the problems, perhaps after reviewing your class notes. Some of the problems are straight from definition of the terms. Not being able to do those problems means you do not have a good grasp of the material.
(1)
The monopoly game uses two six-sided dice and use the sum of the dice to make moves.
The probability of the sum of the dice depends on the sum.
There is only one way to throw a 2 (1+1) or 12(6+6).
There are two ways to throw a 3 (1+2, 2+1) or 11 (6+5,5+6).
There are three ways to throw a 4 (1+3,2+2,3+1) or 10 (4+6,5+5,6+4).
and so on.
The total number of outcomes of throwing two six-sided dice is 36.
This means that throwing a sum of 4 has a probability of 3/36=1/12, and so on.
So to find the probability of landing on Mayfair, first calculate the number of ways to throw a sum of 8, then divide that by 36 to get the probability. This is the probability to land on Mayfair.
The other ones will have an extra step after throwing a six, which gives her a 1/20 probability of whatever the card tells her to do.
So the probability is then
probability=P(sum=6)*P(card)=P(sum=6)/20.
(2)
When A and B are independent, P(A∩B)=P(A)*P(B).
and in general,
P(A∪B)=P(A)+P(B)-P(A∩B)
[use P(A∩B) you calculated in part (A)]
(3)
Two events are mutually exclusive if and only if
P(A∩B)=0
which also means that
P(A∩B)'=1-0=1
See (2) for comments on independence.
(4) there are sixteen possible outcomes.
If you make a list of the sixteen possible outcomes, e.g.
{2+1,2+3,2+5,2+7,4+1,4+3....}
={3,5,7,9,5,7,...}
the answers to each sub-question can be be obtained by checking on the list.
(4)