Question Video: Finding the Probability of Selection without Replacement | Nagwa Question Video: Finding the Probability of Selection without Replacement | Nagwa

Question Video: Finding the Probability of Selection without Replacement Mathematics

A bag contains 8 red balls, 7 green balls, 12 blue balls, 15 orange balls and 7 yellow balls. If two balls are drawn consecutively, without replacement, what is the probability that the first ball is red and the second is blue?

01:34

Video Transcript

A bag contains eight red balls, seven green balls, 12 blue balls, 15 orange balls, and seven yellow balls. If two balls are drawn consecutively without replacement, what is the probability that the first ball is red and the second is blue?

The probability of any event can be written as a fraction, where the numerator is the number of successful outcomes and the denominator is the number of possible outcomes. In this case, there are a total of 49 balls: eight plus seven plus 12 plus 15 plus seven.

We want to work out the probability that the first ball is red and the second one is blue. As there are eight red balls in the bag, the probability that the first ball is red is eight forty-ninths or eight out of 49. There are now 48 balls remaining in the bag of which 12 are blue. Therefore, the probability that the second ball is blue is twelve forty-eighths or 12 out of 48.

As we want the first ball to be red and the second ball to be blue, we need to multiply these two fractions. Eight forty-ninths multiplied by twelve forty-eighths is equal to two forty-ninths. Therefore, the probability that the first ball is red and the second ball is blue is two out of 49 — two forty-ninths.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy