A bag contains 12 colored balls: 4 red, 3 blue, 3 green, and 2 yellow. Four balls are drawn one after another without replacement. What is the probability that the first ball is red, the second ball is blue, the third ball is green, and the fourth ball is yellow?
Options
A
B
C
D
Answer & Analysis
Answer
D
Analysis
Question Analysis
This question involves calculating the probability of four successive dependent events.
The main focus is on applying the concept of dependent - event probability to each draw and using the multiplication rule to find the combined probability of all four events occurring in the specified order.
Key Concept Explanation
For multiple dependent events, each draw changes the composition of the balls in the bag, affecting the probabilities of subsequent draws.
The multiplication rule for dependent events states that the probability of a sequence of dependent events is the product of the probabilities of each event, where the probability of each event is calculated based on the state of the sample space after the previous events have occurred.
Step-by-Step Solution
Calculate the probability of drawing a red ball on the first draw:
There are 4 red balls out of 12 total balls. So, .
Calculate the probability of drawing a blue ball on the second draw given that a red ball was drawn on the first draw:
After removing one red ball, there are 11 balls left, and 3 of them are blue. So, .
Calculate the probability of drawing a green ball on the third draw given that a red ball was drawn first and a blue ball was drawn second:
After removing a red and a blue ball, there are 10 balls left, and 3 of them are green. So, .
Calculate the probability of drawing a yellow ball on the fourth draw given that a red ball was drawn first, a blue ball was drawn second, and a green ball was drawn third:
After removing a red, a blue, and a green ball, there are 9 balls left, and 2 of them are yellow. So,
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