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Worked solution

Use constant probabilities for replacement draws

A box contains blue and white discs. The probability of blue is 1/3 and white is 2/3. Two discs are drawn with replacement. Find the probability that (i) both discs are white, (ii) the first is blue and the second white, (iii) the first is blue or the second white.

CJC · Lecture · Example 9

Define the draw events and keep the probabilities fixed

Let Bi,WiB_i,W_i mean blue and white on draw i. With replacement and independent random drawing, every draw has blue probability 1/3 and white probability 2/3. Follow the Teacher guidance: a tree is unnecessary here because the colour probabilities remain unchanged.

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(i) Multiply the two white probabilities

P(W1∩W2)=(2/3)(2/3)=4/9P(W_1\cap W_2)=(2/3)(2/3)=4/9.

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(ii) Multiply blue-first and white-second probabilities

P(B1∩W2)=(1/3)(2/3)=2/9P(B_1\cap W_2)=(1/3)(2/3)=2/9.

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(iii) Method 1: partition the inclusive union

B1∪W2B_1\cup W_2 is the disjoint union of B1∩W2B_1\cap W_2, B1∩W2′B_1\cap W_2\prime, and B1′∩W2B_1\prime\cap W_2. Since only blue and white exist, these are BW, BB, WW respectively. Their probabilities sum to (1/3)(2/3)+(1/3)(1/3)+(2/3)(2/3)=2/9+1/9+4/9=7/9(1/3)(2/3)+(1/3)(1/3)+(2/3)(2/3)=2/9+1/9+4/9=7/9.

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(iii) Method 2: complement the one excluded order

Neither first-blue nor second-white means first-white and second-blue. Thus P(B1∪W2)=1−P(B1′∩W2′)=1−P(W1∩B2)=1−(2/3)(1/3)=7/9P(B_1\cup W_2)=1-P(B_1\prime\cap W_2\prime)=1-P(W_1\cap B_2)=1-(2/3)(1/3)=7/9. This retains the source’s alternative complement method.

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Common mistakes

“Or” is inclusive. Adding P(first blue) and P(second white) directly counts BW twice. With replacement keeps the composition fixed; independent random draws justify multiplying probabilities.

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