Question

If a ball is drawn from a bag containing 13 red balls numbered 1-13 and 5 white balls numbered 14-18. What is the probability that a. the ball is not even numbered? b. the ball red and even numbered? c. the ball red or even numbered? d. the ball is neither red or even numbered?

Answer 1

Answer:

a. 50%

b. 33%  

c. 17% (I'm assuming the exercise is wrong and it has to say "white" instead of "red", because if not is the same as b.)

d. 67%

Step-by-step explanation:

a. We have a total of 18 balls, 13 are red and 5 are white. They are numbered from 1 to 18. In this case, we don't care about the color of the ball, we just need it to be not even. We have to count how many not even numbers are between 1 and 18, that is 9. So, the chances of getting a ball not even numbered are 9 in 18, that's

$\frac{9}{18}*100=50\%$

b. Now we do care about the color of the ball. The red balls are numbered from 1 to 13, so we have 6 balls even numbered. That makes the chances 6 in 18 (we still have 18 in total), that's

$\frac{6}{18}*100=33\%$

c. (I'm assuming the exercise is wrong and it has to say "white" instead of "red", because if not is the same as b.)

The white balls are numbered from 14 to 18, so we have 3 balls even numbered. That makes the chances 3 in 18,

$\frac{3}{18}*100=17\%$

d. Let's notice that "the ball is neither red or even numbered" is the complement (exactly the opposite) of "the ball is red and even numbered", that means  

100% = Probability (ball red and even numbered) + Probability (ball neither red or even numbered)

So, Probability (ball neither red or even numbered) = 100% - Probability (ball red and even numbered) = 100% - 33% = 67%