Question

A bag contains seven marbles, three red and four green. If three marbles are drawn from the bag at random, what is the probability that all three will be red?
(a) 3/35
(b) 3/7
(c) 1/35
(d) 3/4
(e) 4/7

Answer 1

3/7

Answer 2

all three means first red, second red, third red

Answer 3

no it is not \(\frac{3}{7}\)

Answer 4

Oh yea u have to add them whoops

Answer 5

probability first one is red is \(\frac{3}{7}\)
then the probability the second one is red given that the first one is red is \(\frac{2}{6}\) because there are 2 red ones left, and 6 in the bag
finally the probability that the third is red given the first two are is \(\frac{1}{5}\) because there are now 5 left in the bag and 1 red

Answer 6

Probability of getting the first is 3/7

Answer 7

My mistake didn't fully read question

Answer 8

you get your answer by multiplying these together
i.e.
\[\frac{3}{7}\times \frac{2}{6}\times \frac{1}{5}\]

Answer 9

1/35

Answer 10

that is what i get, yes

Answer 11

thanks

Answer 12

Probability of getting the first is 3/7

Answer 13

yw