Original practice · fully worked
Original variant: delayed shipment count
Let N be the number of delayed shipments in a five-shipment dispatch. Its distribution is P(N=0)=0.64, P(N=1)=0.16, P(N=2)=0.09, P(N=3)=0.06, P(N=4)=0.03, and P(N=5)=0.02. Given that at least one shipment arrives on time, calculate the probability that N is odd.
- A
0.220000 - B
0.224490 - C
0.240000 - D
0.244898 - E
0.775510
Answer and calculation
The condition removes N=5 and has probability 0.98. Within that event, an odd delayed count means N=1 or N=3, with total mass 0.22; dividing gives 0.224490 and choice B.
Setup
Setup
At least one on-time shipment means not all five shipments are delayed.
H={N≤4} Pr(H)=1−Pr(N=5)=1−0.02=0.98 Model
Model
Inside the conditioning event, the possible odd values are one and three; the odd value five has been excluded.
{N odd}∩H={N=1}∪{N=3} Compute
Compute
Add the two eligible point probabilities and normalize by the conditioning probability.
Pr(N odd,H)=0.16+0.06=0.22 Pr(N odd∣H)=0.980.22=0.2244897959… Answer
Answer
The conditional probability of an odd delayed count is approximately 0.224490.
0.224490(B) 