Question Video: Calculating Conditional Probability without Replacement | Nagwa Question Video: Calculating Conditional Probability without Replacement | Nagwa

# Question Video: Calculating Conditional Probability without Replacement Mathematics • Third Year of Secondary School

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A bag contains 22 red balls and 15 black balls. Two balls are drawn at random. Find the probability that the second ball is black given that the first ball is red. Give your answer to three decimal places.

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### Video Transcript

A bag contains 22 red balls and 15 black balls. Two balls are drawn at random. Find the probability that the second ball is black given that the first ball is red. Give your answer to three decimal places.

In this question, we are told that two balls are drawn at random from a bag. One way of representing this is using a tree diagram. We know that the first ball selected can be either red or black. The same is true of the second ball, giving us four possible combinations: red red, red black, black red, or black black. As the two balls are drawn at the same time, we can assume this is done without replacement. This means that we are dealing with dependent events and conditional probability.

Conditional probability can be written using the notation shown: the probability of 𝐵 given 𝐴. In this question, we were asked to find the probability that the second ball is black given that the first ball is red. This is the probability corresponding to the branch highlighted in pink. As the events are dependent, we will firstly need to calculate the probability that the first ball drawn is red.

There are 22 red balls and 15 black balls in the bag. This means there are a total of 37 balls. And the probability that the first ball selected is red is 22 out of 37. Whilst it is not required in this question, we can also add to our tree diagram the probability that the first ball is black. This is equal to 15 out of 37. Let’s now consider how many balls are left in the bag if the first one drawn is red. There are now 21 red balls, and there are still 15 black balls. This is a total of 36, and the probability of selecting a black ball now is 15 out of 36. This is the probability we are looking for. The probability that the second ball is black given that the first ball is red is 15 out of 36.

Whilst we would often leave our answer as a fraction, in this case, we are asked to give our answer to three decimal places. 15 divided by 36 or the simplified fraction five divided by 12 is equal to 0.4166 and so on. We can round this to three decimal places, giving us an answer of 0.417. At this stage, it is worth completing the remainder of the tree diagram. If the first ball selected is red, the probability that the second ball is also red is 21 out of 36, as 21 of the remaining balls are red. If we now assume that the first ball drawn was black, there would be 22 red balls left and only 14 black balls. This means that the probability that the second ball is red given that the first ball is black is 22 out of 36. And the probability that the second ball is black given that the first ball is also black is 14 out of 36.

It is also worth checking at this stage that the three pairs of fractions circled sum to one. We can do this before or after simplifying the fractions.

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