Question

You have a bag of gumballs containing 7 green, 12 blue, 19 yellow, and 3 red gumballs. What is the probability of choosing a red, then a green, then a blue, then a yellow on 4 successive picks if you eat the gumball you pick each time?

Answer 1

Answer:  21/10660

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Work Shown:

7 green + 12 blue + 19 yellow + 3 red = 41 total

A = probability of selecting red first (no replacement)

A = 3/41

B = probability of selecting green second (no replacement)

B = 7/40

C = probability of getting blue third (no replacement)

C = 12/39

D = probability of getting yellow fourth (no replacement)

D = 19/38 = 1/2

E = probability of getting red, green, blue, yellow in that order (no replacement)

E = A*B*C*D

E = (3/41)*(7/40)*(12/39)*(1/2)

E = (3*7*12)/(41*40*39*2)

E = 252/127920

E = 21/10660

When converting to decimal form, this is approximately 0.00196998

There's about a 0.196998% chance of getting this sequence of gumballs if they aren't replaced.