Answer: the first one is 14 and the second one is 5
Step-by-step explanation:
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
Answer: First option is correct.
Step-by-step explanation:
Since we have given that

Now, by factorising , we get

Now, we use the formula i.e.

By using this, we get ,

So,

Hello!
P = 2,1cm + 2,1cm + 1cm + 3,3cm + 1cm => P = 4,2cm + 1cm + 3,3cm + 1cm => P = 5,2cm + 3,3cm + 1cm => P = 8,5cm + 1cm => P = 9,5cm
Answer: D. 9,5cm
Good luck! :)