D=event that chip selected is defective
d=event that chip selected is NOT defective
Four possible scenarios for the first two selections:
P(DDD)=15/100*14/99*13/98=13/4620
P(DdD)=15/100*85/99*14/98=17/924
P(dDD)=85/100*15/99*14/99=17/924
P(ddD)=85/100*84/99*15/98=17/154
Probability of third selection being defective is the sum of all cases,
P(XXD)=P(DDD)+P(DdD)+P(dDD)+P(ddD)
=3/20
Is there a picture I can see
Answer:
Step-by-step explanation:
Time taken in each lap = 4 min
Total laps = 11
Total time = 11 x 4 = 44min
1 /5 = 0 . 2
45 * . 2 = 9
Hope this helped :)