 an urn contains 1 white, 1 green, and 2 red balls. draw 3 balls with replacement. find probability we did not see all the colours Chardonnay Felix 2021-05-21 Answered
an urn contains 1 white, 1 green, and 2 red balls. draw 3 balls with replacement. find probability we did not see all the colours

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Given: Urn contains 1 white, 1 green and 2 red balls
All colors were seen
First we determine the probability of seeing all three colors among the 3 balls.
1 of the 4 balls are white, 1 of the 4 balls are green and 2 of the 4 balls are red. The probability is the number of favorable outcomes divided by the number of possible outcomes:
P(white)= # of favorable outcomes # of possible outcomes=$$\displaystyle\frac{{1}}{{4}}$$
P(green)= # of favorable outcomes # of possible outcomes=$$\displaystyle\frac{{1}}{{4}}$$
P(red)= # of favorable outcomes # of possible outcomes=$$\displaystyle\frac{{2}}{{4}}=\frac{{1}}{{2}}$$
Since the balls are drawn with replacement, the selection of the different balls are independent.
Thus it is then appropriate to use the Multiplication rule for independent events: $$\displaystyle{P}{\left({A}⋂{B}\right)}={P}{\left({A}{\quad\text{and}\quad}{B}\right)}={P}{\left({A}\right)}\cdot{P}{\left({B}\right)}$$.
$$P(\text{all different colors})=P(\text{Red, white and green})=P(red) \cdot P(white) \cdot P(green) =1/2 \cdot 1/4 \cdot 1/4 =1/(2 \cdot 4 \cdot 4) =1/32$$
2.Not all colors were seen Complement rule:
$$\displaystyle{P}{\left({A}^{{c}}\right)}={P}{\left(\neg{A}\right)}={1}-{P}{\left({A}\right)}$$
$$P(\text{not all different colors})=1-P(\text{All different colors})=1-(1/32) =31/32 =0.06875 =96.875%$$