2. A bag contains blue and red marbles. A marble is randomly drawn from the bag
ID: 2909874 • Letter: 2
Question
2. A bag contains blue and red marbles. A marble is randomly drawn from the bag and its colour noted. A second marble is then randomly drawn and its colour is also noted. In each case you may assume that all marbles are equally likely to be drawn from the bag (a) Define the sample space and events for this problem. (b) Suppose the bag contains 6 blue marble and 4 red marbles. 2 marks 1 mark) [1 mark] ii. Compare the two probabilities, explain why what you observe makes intuitive sense. 2 marks i. State the probability that the first marble is blue ii. Compute the probability that the second marble is blue. (c) Suppose instead your bag contains b blue marbles and r red marbles, i. State the probability that the first marble is blue and also state the probability that 2 marks 2 marks [2 marks] (d Suppose instead that your bag contains 9 marbles, and also that the probability that both marbles are the same colour is the same as the probability that you have one of each colour. Find how many blue and red marbles you have. (Note there are two possible 3 marks the second marble is blue ii. Find the probability that both marbles are the same colour ii. What is the probability that you have one marble of each colour? solutions and you should give both)Explanation / Answer
2. (a) S = P{(blue, red), (blue,blue), (red, blue), (red, red)}
(b) i. P(first marble is blue) = 6/(6+4)
= 0.6
ii. P(second marble is blue) = P(blue,blue) + P(red, blue)
= 6/10 x 5/9 + 4/10 x 6/9
= 54/90
= 0.6
iii. The probability of an event will remain the same on any draw because if the event have happened in previous trials, the probability will decrease, whereas if the event did not happen in previous trial, the probability will increase.
c) i. P(first marble is blue) = b/(b+r)
P(second marble is blue) = b/(b+r)
ii. P(both are of same color) = b/(b+r) x (b-1)/(b+r-1) + r/(b+r) x (r-1)/(b+r-1)
= [b(b-1) + r(r-1)]/[(b+r)x(b+r-1)]
iii. P(one marble of each colour) = 2br/[(b+r)(b+r-1)]
d) b+r = 9
[b(b-1) + r(r-1)]/[(b+r)x(b+r-1)] = 2br/[(b+r)(b+r-1)]
b(b-1) + r(r-1) = 2br
b2 - b + r2 - r - 2br = 0
b2 - 2br + r2 - (b+r) = 0
(b-r)2 = (b+r)
(b-r)2 = 9
Ans 1) b - r = 3
b + r = 9
2b = 12
b = 6
r = 3
Ans 2) r - b = 3
b+r = 9
2r = 12
r = 6
b = 3
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