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Answer:

The probability that the second ball is red is 71%

Step-by-step explanation:

Probabilities

We know there are 5 red balls and 2 green balls. Let's analyze what can happen when two balls are drawn in sequence (no reposition).

The first ball can be red (R) or green (G). The probability that it's red is computed by

[tex]\displaystyle P(R)=\frac{5}{7}[/tex]

The probability is's green is computed by

[tex]\displaystyle P(G)=\frac{2}{7}[/tex]

If we have drawn a red ball, there are only 4 of them out of 6 in the urn, so the probability to draw a second red ball is

[tex]\displaystyle P(RR)=\frac{5}{7}\cdot \frac{4}{6}=\frac{10}{21}[/tex]

If we have drawn a green ball, there are still 5 red balls out of 6 in the urn, so the probability to draw a red ball now is

[tex]\displaystyle P(GR)=\frac{2}{7}\cdot \frac{5}{6}=\frac{5}{21}[/tex]

The total probability of the second ball being red is

[tex]\displaystyle P(XR)=\frac{10}{21}+\frac{5}{21}=\frac{5}{7}=0.71[/tex]

The probability that the second ball is red is 71%

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