Question

there are 6 white and 4 orange ping - pong balls in a brown paper bag. two balls are randomly chosen, one after the other without replacement. enter your answers as fractions or as decimals rounded to three decimal places. (the “preview” simply displays your answer in nice mathematical text. it does not mean that your answer is either right or wrong.) a) how many total balls are in the bag at the start? b) what is the probability that the 1st ball is orange? p(1st is orange) = c) what is the probability that the 2nd ball is also orange, given that the 1st ball was orange? p(2nd is orange | 1st is orange) = d) what is the probability that both the 1st and the 2nd balls are orange?

Explanation:

Step1: Calculate total balls

Add white and orange balls. $6 + 4=10$.

Step2: Probability of first - ball being orange

Use probability formula $P(A)=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$. There are 4 orange balls out of 10, so $P(1\text{st is orange})=\frac{4}{10} = 0.4$.

Step3: Probability of second - ball being orange given first is orange

After one orange ball is taken, there are 3 orange balls left and 9 total balls left. So $P(2\text{nd is orange}|1\text{st is orange})=\frac{3}{9}\approx0.333$.

Step4: Probability of both balls being orange

Use the formula for conditional probability $P(A\cap B)=P(A)\times P(B|A)$. Here, $P(\text{both orange})=\frac{4}{10}\times\frac{3}{9}=\frac{12}{90}\approx0.133$.

Answer:

a) 10
b) 0.4
c) 0.333
d) 0.133