Question

a bag contains 4 blue and 3 yellow marbles. the marbles are removed from the bag one at a time without replacement. what is the probability that the fifth marble removed is yellow? express your answer as a common fraction.

Answer 1

can you please help me...

Answer 2

yes i can... :p
wait

Answer 3

P(B)*P(B)*P(B)*P(B)*P(Y) = 4/7 * 3/6 * 2/5 * 1/4 * 3/3 = ....

Answer 4

hmmm

Answer 5

what is P(B)

Answer 6

there are other possibilities besides
B B B B Y

Answer 7

probabilty of event BLUE
no, sattelite... the question is the fifth is yellow, so the 1st, 2nd, 3rd, 4th are blue

Answer 8

@RadEn it didn't say the 5th one is THE FIRST yellow.

Answer 9

so what is the answer

Answer 10

??

Answer 11

there is implicit message, there @r0n4ld

Answer 12

let y be the number of yellow balls drawn on the first four. clearly y = 0, 1, 2.

Answer 13

i got, 24/840 = 1/35
do u have choices

Answer 14

since there are 4 balls drawn before the 5th which is yellow, the total number of arrangement of y yellow balls and (4-y) blue balls is \[\frac{4!}{y!(4-y)!}\]

Answer 15

sum these values from y= 0 to 2...: = 11
my answer is \[\large\frac{11}{9\cdot8\cdot7\cdot6\cdot5}\]

Answer 16

i think that is much snappier than what i had in mind

no yellow balls, 5th one is yellow
one yellow ball, 5th one is yellow
two yellow balls, 5th one is yellow

compute these probabilities and add them up is what i had in mind