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
475.82
is the answr i think
The second and last one.
The second one multiplies the cost of each type of ticket by two, therefore saying that you wanted to buy two of each type of ticket.
The last one multiplies the cost of a single ticket of all three types and multiplies it by 2.
Answer:
The answer is 9 + 21b
Step-by-step explanation:
Simplify the expression.
~Hoped this helped~