A bag contains 5 red balls and 5 blue balls. If you draw two balls without replacement, what is the probability that the second ball is red, given that the first ball was red?

Question

BlackTom AI · Accepted Answer

Consider the scenario step by step, focusing on how the composition of the bag changes after the first draw.
Option a: 5/10. This would imply no change in the counts after the first red draw, treating the second draw as if the total and red count remained the same. Since we removed one red ball, both the numerator and denominator should reflect the updated totals, so this is incorrect.
Option b: 4/9. After drawing a red ball first, the bag now contains 4 red and 5 blue balls, for a total of 9 balls. The probability that the next ball is red is the number of remaining red balls divided by the new total: 4/9. This matches the updated composition, so this is correct.
Option c: 5/9. This would be the probability if the red count remained at 5 while the total reduced to 9, which cannot happen since one red ball was removed. It miscounts both the remaining red and total, making it incorrect.
Option d: 4/10. This would correspond to 4 red out of 10 total, implying no reduction in the total number of balls, which is not the case after the first draw in a without-replacement scenario. Therefore this is not right.

Statistics and Probability

A bag contains 5 red balls and 5 blue balls. If you draw two balls without replacement, what is the probability that the second ball is red, given that the first ball was red?

View Explanation

Log in for full answers

Similar Questions

Question at position 7 CAEBD

Given that the first die shows a 3, what is the probability that S is an even number?

Given that S is 7, what is the probability that the first die showed 4 dots?

A survey found that 60% of households own a dog (D), 40% own a cat (C), and 20% own both a dog and a cat (D∩C). What is the probability that a randomly selected household owns a dog, given that they own a cat?

If for two events [math: A] and [math: B] , [math: Pr(A)=920] Pr(A)= \frac{9}{20} , [math: Pr(B)=12] Pr(B)= \frac{1}{2} , and [math: Pr(A∩B)=15] Pr(A\cap B)= \frac{1}{5} , then [math: P(A|B)] is equal to :

A box contains 10 red, 5 green, and 10 blue marbles. A marble is chosen at random from the box, and it is noted that it is not one of the blue marbles. What is the probability that it is green?

Based on weather data collected in Racine, Wisconsin, on Christmas Day, the weather had the following distribution: Event Relative Frequency Clear & dry 0.20 Cloudy & dry 0.30 Rain 0.40 Snow 0.10 Supposing next Christmas is dry, determine the probability that it will also be cloudy.

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler