3.(8 points)Consider the following experiment:Experiment: A bag contains 3 red balls, 4 blue balls, 4 green balls and 3yellow balls. Two balls are drawn from the bag without replacing the ballafter the first draw.(a)(2 points)Specify the sample space (set of outcomes) and probability distributionof the outcomes.(b)(3 points)If you draw two balls from the bag without replacing them, what isthe probability of drawing two red balls

3 / 91

Step-by-step explanation:

Given :

Red balls = 3

Blue balls = 4

Green balls = 4

Yellow balls = 3

Total number of balls = (3+4+4+3) = 14

Sample space :

RR, BB, GG, YY, RB, BR, BG, GB, YG, GY, YR, RY, BY, YB, GR, RG

Probability of drawing two red balls without replacement :

1st draw :

P(red) = number of red balls / total number of balls

P(red) = 3/14

2nd draw :

P(red) = 2 / 13

Hence,

P(RR) = 3/14 * 2/13 = 6 / 182 = 3 / 91

the answer is:

the answer is:

to solve the problem, we need to remember the quotient of power property, it's defined by the following relation:

if we have a quotienf of powers that have the same base, we need to keep the same base and subtract the exponent of the denominator power to the exponent of the numerator power.

so, we are given the expression:

then, calculating we have:

hence, the answer is:

have a nice day!

answer: 1475

step-by-step explanation:

becuase you have to know what they are asking you