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  • What is the probability of drawing a red ball on the first draw and a blue ball on the second draw without replacement, from a bag containing 5 red balls and 3 blue balls?<div class='qna-option

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

Option: 1

\frac{15}{56}


Option: 2

\frac{15}{64}


Option: 3

\frac{5}{56}


Option: 4

\frac{5}{64}


Answers (1)

best_answer

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

Two events are independent if the occurrence of one event does not affect the occurrence of the other event. 

Let's denote the events as follows: 

R: the event that the first ball selected is red 

B: the event that the second ball selected is blue 

We need to find the probability of the intersection of these two events, P(R and B). 

To do this, we can use the formula:

{P(R \ \text{and} \ B)=P(R)\times P(B/R)}

Where P(R) is the probability of selecting a red ball on the first draw, and P(B|R) is the probability of selecting a blue ball on the second draw given that the first ball drawn was red. 

First, let's calculate P(R):

\\{P(R)=\frac{\text{Number of red balls}}{\text{Total number of red balls on the first draw}}}\\ \\\Rightarrow {P(R)=\frac{5}{8}}

Next, let's calculate P(B|R): 

Since we are drawing without replacement, the total number of balls on the second draw will be 7 (since one red ball has already been drawn).

\\{P(B/R)=\frac{\text{Number of blue balls remaining}}{\text{Total number of balls remaining}}}\\ \\\Rightarrow {P(B/R)=\frac{3}{7}}

Now we can use the formula to find the probability of the intersection:

{ P(R \ \text{and} \ B)=P(R)\times P(B/R)}

\\{P(R \ \text{and} \ B)=\frac{5}{8}\times \frac{3}{7}}\\ \\\Rightarrow { P(R \ \text{and} \ B)=\frac{15}{56}}

Therefore, the probability that the first ball selected is red and the second ball selected is blue is \frac{15}{56}.

Posted by

Irshad Anwar

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