WEBVTT
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A bag contains 22 red balls and nine green balls.
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One red ball is removed from the bag.
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And then a ball is drawn at random.
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Find the probability that the drawn ball is red.
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Initially, there are 22 red balls and nine green balls in the bag.
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However, one of the red balls is removed.
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This means that we now have 21 red balls and nine green balls remaining in the bag.
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In order to calculate the probability, we need to divide the number of successful outcomes, in this case the red balls, by the number of possible outcomes, the total number of balls.
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As there are 21 red balls remaining in the bag, our numerator, or top number, of our fraction will be 21.
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The total number of balls remaining in the bag is 30.
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21 red plus nine green equals 30.
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Therefore, we can say that the probability of drawing a red ball is 21 out of 30, or 21 divided by 30.
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This fraction can be simplified by dividing the top, the numerator, and the bottom, the denominator, by three.
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21 divided by three is seven.
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30 divided by three is 10.
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Therefore, the probability of drawing a red ball is seven tenths, or seven over 10.
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We could also write this answer as a decimal, 0.7, or a percentage, 70 percent.
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There is a 70 percent chance of drawing a red ball from the bag.