Hi Guys, how would I do all of question 18

Can't read your attachment...please post the problem yourself.
Also, please show where you're stuck.
 
Denis said:
Can't read your attachment...please post the problem yourself.
Also, please show where you're stuck.
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Denis, just click on the image, then use the pointing device to magnify.
 
pka said:
These are extremely easy questions. What have you tried? Where are you stopped.
Please give complete answer so we can help.
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Sorry I'm only a Yr 11 student, have I come to the wrong place? I just have a hard time with probability.
 
OK. One way of looking at this question is using probability tree diagrams. Do you know what they are?
 
In order to help best, we need to know what you have learned, so we can use that to help. (See this summary of our posting guidelines.)

Do you know how to work with "and" probabilities?

Since the disc is being replaced before the next draw, the selections are independent. Do you know the importance of that?

Finally, what is the probability that Marnie's disc is red?

Then see if you can put those together. And let us know if you have not learned any of the ideas I've asked about.
 
A bag contains 10 red, 7 blue, and 3 green disks.

The probability of choosing a red disk at random is 10/20= 1/2. The probability that all three pick a red disk is (1/2)^3= 1/8..

Reggie takes 3 disk out of the bag, not returning them to the bag.
What is the probability that all three disks are of different color.
The probability the first disk is red is 10/20= 1/2. There are then 9 red, 7 blue, and 3 green disk. The probability the second disk is blue is 7/19. There are the 9 red, 6 blue, and 3 green. The probability the third disk is green is 3/18= 1/6. The probability of "red, blue, green" in that order, is (1/2)(7/19)(1/6)= 7/228.

The probability the first disk is red is 10/20= 1/2. There are then 9 red, 7 blue, and 3 green disk. The probability the second disk is green is 3/19. There are the 9 red, 7 blue, and 2 green. The probability the third disk is blue is 7/18. The probability of "red, blue, green" in that order, is (1/2)(3/19)(7/19)= 7/228.

The probability the first disk is blue is 7/20. There are then 10 red, 6 blue, and 3 green disk. The probability the second disk is red is 10/19. There are the 9 red, 6 blue, and 3 green. The probability the third disk is green is 3/18= 1/6. The probability of "red, blue, green" in that order, is (7/20)(10/19)(1/6)= 7/228.

Get the point? There are 3!= 6 different orders of "red, green, and blue" in any order so the probability of "all three different colors" is 6(7/228)= 7/38.
 
HallsofIvy said:
A bag contains 10 red, 7 blue, and 3 green disks.
The probability the first disk is blue is 7/20. There are then 10 red, 6 blue, and 3 green disk. The probability the second disk is red is 10/19. There are the 9 red, 6 blue, and 3 green. The probability the third disk is green is 3/18= 1/6. The probability of "red, blue, green" in that order, is (7/20)(10/19)(1/6)= 7/228.
Get the point? There are 3!= 6 different orders of "red, green, and blue" in any order so the probability of "all three different colors" is 6(7/228)= 7/38.
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Given a bag of 20 discs (10 red, 7 blue, & 4 green disks.)
Part b. of #18 says That all 20 discs are in a bag.
Reggie takes three discs at random, one after the other.
After he takes a disc, he does not return it to the bag.
Reggie's first disc is blue.
Work out the probability that all three discs are different colours.
Now I copied the question exactly, even preserving the British spelling.

Now I take it that the question is asking for the probability:
\(\displaystyle \mathscr{P}(B_1\,G_2\,R_3)\) or \(\displaystyle \mathscr{P}(B_1\,R_2\,G_3)\)

NOTE that I do not take that the author expects conditional probability to be used.
In that case I think the intended answer is \(\displaystyle 2\cdot\left(\dfrac{7}{20}\right)\cdot\left(\dfrac{10}{19}\right)\cdot\left(\dfrac{4}{18}\right)\).

In asking SpoderCactus to respond I had hoped to get a supplied answer to compare it with my answer.
However, had I edited this text, I have insisted the answer should have been a conditional.
 
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