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Probability of choosing 1 blue and other red ball from a bag containing 1 red , 1 blue and 3 green balls. Without replacement

Question

Probability of choosing 1 blue and other red ball from a bag containing 1 red , 1 blue and 3 green balls. Without replacement

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Solution

Sure, here are the steps to solve this problem:

Step 1: Identify the total number of balls in the bag. In this case, there are 1 red, 1 blue, and 3 green balls, which makes a total of 5 balls.

Step 2: Calculate the probability of drawing a blue ball first. This is the number of blue balls (1) divided by the total number of balls (5). So, the probability of drawing a blue ball first is 1/5.

Step 3: Since we are drawing without replacement, we now have 4 balls left in the bag. The probability of drawing a red ball next is the number of red balls (1) divided by the new total number of balls (4). So, the probability of drawing a red ball next is 1/4.

Step 4: The probability of both events happening (drawing a blue ball first and a red ball second) is the product of the probabilities of each event. So, multiply the two probabilities together: (1/5) * (1/4) = 1/20.

So, the probability of drawing one blue ball and one red ball from the bag, in that order, without replacement, is 1/20.

This problem has been solved

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