A bag contains 1 red disc, 2 blue discs and 3 green discs.
Ben chooses ta disc at random from the bag. He notes its colour and replaces it. Then he chooses another disc at random from the bag and notes its colour.
1) Calculate the prob that both discs are the same colour?
Is the answer: (1/6)squared + (2/6)squared + (3/6)squared = 7/18
2) Calculate the prob that neither disc is red?
Is the answer: 1 - (1/6)squared = 35/36 ? I fear I might be wrong on this one. But not show why?
Nov 20th 2010, 12:26 PM
The first one is correct because the outcomes that you have calculated are mutually exclusive. The second one is correct as well, as neither suggest both are not red. Why did you feel the answer was not correct?
Nov 20th 2010, 12:29 PM
Because it gives me 97.2 % and I can't understand if there are 6 discs in a bag it would take me 97.2 % (chance) for me to neither get reds
Nov 20th 2010, 01:08 PM
I'm sorry I misread the problem. Your answer to Part B is incorrect. Let's look at what it is is asking - "What is the probability that neither disc is red". This means we want to sample (with replacement) two discs, and we want BOTH of them to NOT be red. I see where you were going with 1-P(RR), but that still leaves BR, RB, BG, RG as options - which we don't want. Neither means - none of the discs can be red. So we need to correct for that. What you have now, 1-P(RR) is the probability that BOTH discs are not red, which is not the same as NEITHER.
Nov 20th 2010, 01:10 PM
Yes exactly, that's what I thought. Still a little stuck...
Nov 20th 2010, 01:13 PM
What part are you stuck on?
Nov 20th 2010, 01:19 PM
well lets see...
That leaves us with this prob = (2/6)squared + (2/6 x 3/6) + (3/6 x 2/6) + (3/6)squared is this correct?