Suppose that 11 green balls and 14 purple balls are placed in an urn. two balls are then drawn in succession. what is the probability that both balls drawn have the same color if the first ball is replaced before the second is drawn?

The probability that both balls drawn have the same color if the first ball is replaced before the second is drawn is;

P(both draws same color) = 50.72%

We are given that;

Number of green balls = 11

Number of purple balls = 14

Total number of balls = 11 + 14

Total number of balls = 25 balls

Now, probability of drawing a green ball first is;

P(G1) = 11/25

Now,we are told that the first selection is replaced

Thus, probability of drawing a green ball at second pick is;

P(G2) = 11/25

Thus, probability of drawing 2 consecutive green balls is;

P(GG) = P(G1) × P(G2)

P(GG) = (11/25)²

P(GG) = 0.1936

Similarly, probability of drawing a purple ball first is;

P(P1) = 14/25

Now,we are told that the first selection is replaced

Thus, probability of drawing a purple ball at second pick is;

P(P2) = 14/25

Thus, probability of drawing 2 consecutive purple balls is;

P(PP) = P(P1) × P(P2)

P(PP) = (14/25)²

P(PP) = 0.3136

Thus, probability that both balls drawn have the same color if the first ball is replaced before the second is drawn is;

P(both draws same color) = P(GG) + P(PP)

P(both draws same color) = 0.1936 + 0.3136

P(both draws same color) = 0.5072 or 50.72%

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step-by-step explanation:

the answer is 1030 it's very simple math.