Quick help in easy-ish probability

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In summary, the conversation involved finding the probability of prize winners A and B both receiving cars, either A or C or both receiving cars, A receiving a car and B receiving either a car or a bicycle, and A receiving a car given that B gets either a car or a bicycle. The correct calculations were provided for each scenario.
  • #1
rock.freak667
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Homework Statement



Here is the basic info. There are 24 prizes(4 cars,8 bicycles and 12 watches) to be given to 24 prize winners. A,B,C are 3 of the prize winners.
Find
(i)The probability that both A and B get cars given that C gets a car.
(ii)The probability that either A or C or both gets a car
(iii)The probability that A gets a car and B gets either a car of a bicycle
(iv)The probability that A gets a car given then B gets either a car or a bicycle.

Homework Equations


The Attempt at a Solution



(i)P([itex]A_c&B_c|C_c[/itex])=[tex]\frac{\frac{3}{23}*\frac{2}{22}}{\frac{4}{24}}[/tex]

(ii)simply I got this one by finding when they both did not get cars and then 1 minus that value.

(iii)[itex]\frac{4}{24}*\frac{3}{23}+\frac{4}{24}*\frac{8}{23}[/itex]

(iv)This part confused me as I got a way wrong answer.(ANS=[itex]\frac{11}{69}[/itex], which I got something way smaller than that)
 
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  • #2
(iv) Does the sentence read "The probability that A gets a car given that B gets either a car or a bicycle."

EDIT: correct method follows.
 
Last edited:
  • #3
Shooting star said:
(iv) Does the sentence read "The probability that A gets a car given that B gets either a car or a bicycle."

If that is the case, then P(B gets car)*P(A gets car) + P(B gets bicycle)*P(A gets car) = 4/24*3/23 + 8/24*4/23 = 4*11/(24*23) = 11/(6*23) = (1/2)*11/69.

Aren't you supposed to use
[tex]P(A|B)=\frac{P(A\cap B)}{P(B)}[/tex]
 
  • #4
All right. What is P(B) in that case? More importantly, what would be P(A_int_B)?
 
  • #5
Shooting star said:
All right. What is P(B) in that case? More importantly, what would be P(A_int_B)?


B should be the probability of getting a car or a bicycle which is 4/24+8/24.

and so P(A_int_B)=4/24*(4/24+8/24)
 
  • #6
rock.freak667 said:
and so P(A_int_B)=4/24*(4/24+8/24)

Shouldn't it be P(A_int_B)=4/24*(4/23+8/23)? Take a sec...
 
  • #7
Which gives you again your old answer...

Do you know the correct result?
 
  • #8
Shooting star said:
Shouldn't it be P(A_int_B)=4/24*(4/23+8/23)? Take a sec...
oh it is 23 because A gets a prize and there are 23 left...stupid me.

Shooting star said:
Which gives you again your old answer...

Do you know the correct result?

the answer is 11/69
 

FAQ: Quick help in easy-ish probability

How can I quickly calculate the probability of an event?

To calculate the probability of an event, you need to divide the number of favorable outcomes by the total number of possible outcomes. For example, if you are rolling a six-sided die and want to know the probability of rolling a 3, you would divide 1 (the number of favorable outcomes) by 6 (the total number of possible outcomes), which gives you a probability of 1/6 or approximately 0.1667.

What is the difference between independent and dependent events in probability?

Independent events are events that do not affect each other's outcome. For example, if you are flipping a coin and rolling a die, the outcome of one event does not affect the outcome of the other. Dependent events, on the other hand, are events where the outcome of one event does affect the outcome of the other. For instance, if you are drawing cards from a deck without replacing them, the probability of drawing a certain card will change with each draw.

How do I use the multiplication rule in probability?

The multiplication rule states that to find the probability of two independent events occurring together, you need to multiply their individual probabilities. For example, if you are flipping a coin twice, the probability of getting heads both times would be 1/2 * 1/2, which equals 1/4 or 0.25.

How can I use the addition rule in probability?

The addition rule is used to find the probability of either one of two or more events occurring. It states that you need to add the individual probabilities of the events together, as long as the events are mutually exclusive (meaning they cannot occur at the same time). For example, if you are rolling a six-sided die and want to know the probability of rolling either a 2 or a 4, you would add 1/6 (the probability of rolling a 2) and 1/6 (the probability of rolling a 4), which equals 2/6 or approximately 0.3333.

What is the difference between permutations and combinations in probability?

Permutations are the different ways in which a set of objects can be arranged, while combinations are the different ways in which a selection of objects can be chosen from a larger set. In probability, permutations are used when the order of the events matter, while combinations are used when the order does not matter. For example, if you are choosing a president, vice president, and treasurer from a group of 10 people, you would use permutations because the order of the positions matters. But if you are choosing a team of 3 people to work on a project, you would use combinations because the order of the team members does not matter.

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