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# Any help would be appreciated

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The probability of drawing two red candies without replacement is 13/35 ,  and the probability of drawing one red candy is 2/5 .

What is the probability of drawing a second red candy, given that the first candy is red?

Apr 20, 2018

#1
+985
+2

Alright, here we go:

To draw two red candies without replacement is 13/35, but since drawing the first red candy is 2/5, drawing the second one, must be:

$$\frac{13}{35}\div\frac{2}{5}=\frac{13}{14}$$

.
Apr 21, 2018

#1
+985
+2

Alright, here we go:

To draw two red candies without replacement is 13/35, but since drawing the first red candy is 2/5, drawing the second one, must be:

$$\frac{13}{35}\div\frac{2}{5}=\frac{13}{14}$$

GYanggg Apr 21, 2018
#2
+101768
+2

The probability of drawing two red candies without replacement is 13/35 ,  and the probability of drawing one red candy is 2/5 .

What is the probability of drawing a second red candy, given that the first candy is red?

$$\frac{2}{5 }x=\frac{13}{35}\\ x=\frac{13}{35}\div \frac{2}{5 }\\ x=\frac{13}{35}\times \frac{5}{2 }\\ x=\frac{13}{14}$$

.
Apr 21, 2018