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Worked solution

Keep the two dice ordered when checking event independence

Two fair dice, one white and one red, are thrown. Event A is that the white score is prime; event B is that the sum of the scores is 9.

(i) Find P(A∣B)P(A\mid B) and hence determine whether A and B are independent. marks

(ii) Event C is that the red score is a perfect square. Find P(B′∩C′)P(B'\cap C'). marks

[2014/RVHS/II/6]

CJC · Review problems · Question 6

Answer

(i) 1/21/2, and A,B are independent; (ii) 7/127/12.

(i) Identify the favourable ordered pairs

Write each pair as(white, red). The white primes are 2,3,5, so P(A)=3/6=1/2P(A)=3/6=1/2. Sum 9 occurs for (3,6),(4,5),(5,4),(6,3), hence P(B)=4/36=1/9P(B)=4/36=1/9. The intersection A∩B consists of (3,6),(5,4), so P(A∩B)=2/36=1/18P(A\cap B)=2/36=1/18.

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(i) Conditional-probability test

P(A∣B)=P(A∩B)/P(B)=(1/18)/(1/9)=1/2=P(A)P(A\mid B)=P(A\cap B)/P(B)=(1/18)/(1/9)=1/2=P(A), so A and B are independent.

How did this part go?

(i) Source alternative: product test

P(A)P(B)=(1/2)(1/9)=1/18=P(A∩B)P(A)P(B)=(1/2)(1/9)=1/18=P(A\cap B), establishing the same independence.

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(ii) Complement the union

The red-die squares are 1 and 4, so P(C)=2/6P(C)=2/6. Among the sum 9 pairs, only (5,4) also has a square red score, giving P(B∩C)=1/36P(B\cap C)=1/36. Therefore P(B∪C)=4/36+2/6−1/36=5/12P(B\cup C)=4/36+2/6-1/36=5/12, and P(B′∩C′)=1−P(B∪C)=7/12P(B'\cap C')=1-P(B\cup C)=7/12. The source diagram shades the outside of both circles.

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Expand diagram

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Two overlapping event circles B and C, both white, inside a blue sample-space rectangle. Blue is the complement of their union, B-complement intersection C-complement, reproducing the source outside-both shading.
Two overlapping event circles B and C, both white, inside a blue sample-space rectangle. Blue is the complement of their union, B-complement intersection C-complement, reproducing the source outside-both shading.
Two overlapping event circles B and C, both white, inside a blue sample-space rectangle. Blue is the complement of their union, B-complement intersection C-complement, reproducing the source outside-both shading.

Common mistakes

1 is a perfect square but not a prime. Do not interchange the white-prime and red-square conditions.

Solution notes

Every printed case, both independence methods, union calculation, outside-both source diagram and marks retained.

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