Question

A goblet contains 333 red balls, 222 green balls, and 666 blue balls. if we choose a ball, then another ball without putting the first one back in the goblet, what is the probability that the first ball will be red and the second will be blue?

Answer 1

Step-by-step explanation:

4

89

69849

55

Answer 2

Answer:

$\frac{9}{55}$

Step-by-step explanation:

A goblet contains 3 red balls, 2 green balls and 6 blue balls.

Total balls = 3 + 2 + 6 = 11 balls

we choose a ball and then another ball without putting the first one back.

The probability that the first ball will be red = $P_{1}$ = $\frac{3}{11}$

Second event is done without replacing the first ball so total ball will be 10

The probability that the second ball will be blue = $P_{2}$ = $\frac{6}{10}$ = $\frac{3}{5}$

P = $\frac{3}{11}$ × $\frac{3}{5}$ = $\frac{9}{55}$

The probability that the first ball will be red and the second ball will be blue is $\frac{9}{55}$