Question

Help! Anyone can you explain?

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

Answer: 9/55

P(1st = red and 2nd = blue)

= (3/11) x (6/10)

= 18/110

= 9/55

To find the probability of something happening,

= (number of desired outcomes) / (total number of outcomes)

If you are finding the probability of more than one thing happening at the same time, you multiply the probability of both things happening together

Answer 2

`\texttt{Probability that the first ball will be red and the second will be blue = }(9)/(55)`

Explanation:

Total number of balls = 3 + 2 + 6 = 11

Probability is the ratio of number of favorable outcome to total number of outcomes.

`\texttt{Probability that the first ball will be red = }\frac{\texttt{Total number of red balls}}{\texttt{Total number of balls}}\\\\\texttt{Probability that the first ball will be red = }(3)/(11)`

Now we have 10 balls in which 6 are blue.

`\texttt{Probability that the second ball will be blue = }\frac{\texttt{Total number of blue balls}}{\texttt{Total number of balls}}\\\\\texttt{Probability that the first ball will be red = }(6)/(10)=(3)/(5)`

`\texttt{Probability that the first ball will be red and the second will be blue = }(3)/(11)* (3)/(5)\\\\\texttt{Probability that the first ball will be red and the second will be blue = }(9)/(55)`