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  • What is the probability of drawing a red ball on the first draw and a green ball on the second draw without replacement, from a box containing 5 red balls, 3 green balls, and 2 blue balls?<div

What is the probability of drawing a red ball on the first draw and a green ball on the second draw without replacement, from a box containing 5 red balls, 3 green balls, and 2 blue balls?

Option: 1

\frac{1}{2}


Option: 2

\frac{1}{4}


Option: 3

\frac{1}{6}


Option: 4

\frac{1}{8}


Answers (1)

best_answer

To solve this problem, we need to understand the concept of independent events of probability. 

Two events A and B are independent if the occurrence of A does not affect the occurrence of B. 

In this case, the probability of drawing a red ball on the first draw is   \frac{5}{10}  (since there are 5 red balls out of 10 total balls). 

After one red ball is drawn, there are 4 red balls and 9 total balls left in the box. 

Now, the probability of drawing a green ball on the second draw, given that a red ball was

drawn on the first draw, is  \frac{3}{9} (since there are 3 green balls left out of 9 total balls). 

To find the probability of both events occurring (drawing a red ball on the first draw and a green ball on the second draw), we multiply the probabilities: 

\text{P(red first, green second)}=\text{P(red first)}\times \text{P(green second/ red first)}

\\\Rightarrow \text{P(red first, green second)}=\frac{5}{10}\times \frac{3}{9}\\ \\\Rightarrow \text{P(red first, green second)}=\frac{1}{6}

Therefore, the probability of drawing a red ball on the first draw and a green ball on the second draw is \frac{1}{6}.

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Pankaj

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