Probability involving Combinations and Permutations (1 Viewer)

[Smilebuffalo]

3.a) An urn contains 3 white balls, 4 red balls and 5 black balls. Three balls are drawn at random. What is the probability that they are different colours? (answer = 3/11)

22. Urn A contains 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. From urn A two balls are selected at random and placed in urn B. From urn B two balls are then selected at random. What is the probability that exactly one of these two balls is white? (answer = 128/225)

27.c) Six people, of whom A and B are two, arrange themselves at random in a row. What is the probability that there are at least three people between A and B? (answer = 2/5)

28. The digits 1,2,3,4,5,6 are used to form numbers which contain 2 or more digits. (The same digit cannot be used more than once in a number). What proportion of such numbers are even numbers? (answer = 1/2)

 

[daryl-d]

the first one

P(E)= 3C1 X 4C1 X 5C1 / 12C3

for the second one consider the three situations ie when two white balls are drawn from A, or one white and one black ball is drawn or 2 black balls are drawn... find the individual probablities when these are all drawn from B add these up and that should hopefully get the answer

 

[ninetypercent]

28. The digits 1,2,3,4,5,6 are used to form numbers which contain 2 or more digits. (The same digit cannot be used more than once in a number). What proportion of such numbers are even numbers? (answer = 1/2)

the number must end in an even number. there are 6 numbers, 3 even. chances are 1/2

 

[Smilebuffalo]

ahh okay i understand Q3 and 28, but i still dont' know how to do question 22. Also does anyone know how to do Q27c)?

 

[ninetypercent]

22. Urn A contains 6 white and 4 black balls. Urn B contains 2 white and 2 black balls. From urn A two balls are selected at random and placed in urn B. From urn B two balls are then selected at random. What is the probability that exactly one of these two balls is white? (answer = 128/225)

say that the balls you select from Urn A are WW => 6C2/10C2
chances of selecting WB from Urn B => (4C1 x 2C1)/6C2
probability = 6C2/10C2 x (4C1 x 2C1)/6C2 = 8/45

say that the balls u select from Urn A are WB => (4C1 x 6C1)/10C2
chances of selecting WB from Urn B => (3C1 x 3C1) /6C2
probability = (4C1 x 6C1)/10C2 x (3C1 x 3C1)/6C2 = 8/25

say that the balls u select from Urn A are BB => 4C2/10C2
chances of selecting WW from Urn B => (2C1 x 4C1)/6C2
probability = 16/225

total = 16/225 + 8/25 + 8/45 = 128/225

 

[ninetypercent]

27.c) Six people, of whom A and B are two, arrange themselves at random in a row. What is the probability that there are at least three people between A and B? (answer = 2/5)

consider that there can be 3 or 4 people between A and B. then consider the cases for when A is in first position and second position.

 

[Smilebuffalo]

27.c) Six people, of whom A and B are two, arrange themselves at random in a row. What is the probability that there are at least three people between A and B? (answer = 2/5)

consider that there can be 3 or 4 people between A and B. then consider the cases for when A is in first position and second position.

Bah I can't get it. For some reason i keep getting 1/5 instead of 2/5. Sorry to bother you again, but could you post a worked solution to the question?

 

[ninetypercent]

so you can have A _ _ _ B _
A _ _ _ _ B
_ A _ _ _ B

A and B can alternate in each case, and there are 3 cases. the remaining people can sit in 4! places.

so 3 x 4! x 2 = 144

no. of ways without restriction = 6! = 720

probability = 1/5

I didn't get it too...

 

[lolokay]

did you consider that maybe the answer they gave is incorrect?