Answer:
Probability ball: white - P(W)P(W);
Probability ball: even - P(E)P(E);
Probability ball: white and even - P(W&E).
Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).
(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)−0P(WorE)=P(W)+P(E)−0. Not sufficient
(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)−P(E)=0.2P(W)−P(E)=0.2, multiple values are possible for P(W)P(W) and P(E)P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)P(WorE).
(1)+(2) P(W&E)=0 and P(W)−P(E)=0.2P(W)−P(E)=0.2 --> P(WorE)=2P(E)+0.2P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4P(E)=0.4 (10 even balls) then P(WorE)=1P(WorE)=1 BUT if P(E)=0.2P(E)=0.2 (5 even balls) then P(WorE)=0.6P(WorE)=0.6. Not sufficient.
Answer: E.
Answer:
2000
Step-by-step explanation:
just muliply
The answer is 5
Hope this helps!
Answer:
1.79438504
Step-by-step explanation:
This is the answer for your question above
