Class 12 Probability · Case Study4 Marksprobability_distributions
Class 12 Maths · Probability · Case Study #129
An octagonal prism has8 rectangular side faces labeled1 to8. It is rolled along these faces, and X is the bottom label. Its probability distribution is: X=1,2,3,4,5,6,7,8; P(X)=p,2p,2p,p,2p,p²,2p²,7p²+p. This is a specified weighted model, not an assumption of equally likely faces.
Part (1) 1 Marks
Find p.
Answer: p=1/10
- Sum the masses: 10p²+9p=1, so (10p−1)(p+1)=0.
- The root −1 gives a negative mass; therefore p=1/10.
Part (2) 1 Marks
Find P(X>6).
Answer: 19/100
- P(X>6)=P(7)+P(8)=2p²+7p²+p=9p²+p.
- At p=1/10 this equals9/100+1/10=19/100.
Part (3) 2 Marks
Find P(X=3m for some positive integer m).
Answer: 21/100
- The admissible multiples of3 in {1,…,8} are3 and6.
- Their disjoint probabilities sum to2p+p². At p=1/10 this is1/5+1/100.
- Thus the probability is21/100.
Part (4) 2 Marks
Find E(X).
Answer: 203/50=4.06
- Use E(X)=ΣxP(X=x).
- The weighted sum is p+4p+6p+4p+10p+6p²+14p²+8(7p²+p)=33p+76p².
- At p=1/10, E(X)=33/10+76/100=203/50=4.06.