# probability marble question

• May 17th 2006, 08:13 AM
aptiva
probability marble question
hi
I am having trouble answering this question from my text
can anyone help me?

A bag contains 8 blue marbles, 7 red marbles, and 5 green marbles

a) If you randomly select 3 marbles with replacement, what is the probability of drawing a red, a green and a blue marble in that order.

b) If you randomly select 3 marbles without replacement, what is the probability of drawing a red, a green, and a blue marble in that order.

c) If you randomly select marbles without replacement, what is the probability that at least 2 of them will be red.

a)0.035
b)0.041
c)0.2702
• May 17th 2006, 01:47 PM
gogo08
I can give an explanation for the first two. The last one will take a little longer.

First find your sample size 8 + 7 + 5= 20

A. You want red, green and blue so you multiply 7/20 * 5/20 * 8/20= 280/8000 which divided out is .032

B. Since you can't replace your probability changes but you still multiply 7/20 * 5/19 * 8/18 (you need to subtract one each time in the demoninator since your sample size changes after each marble is chosen.) multiply that out and you get 280/6840 which divides out to be .0408
• May 17th 2006, 01:47 PM
Rich B.
Hi:

In keeping with convention, let the terminology define P(u) as "the probability that event-u occurs". Then, we could pose the opening question as:

#1) Determine P(R,G,B) for permutation R,G,B taken with replacement on {7R, 5G, 8B}.

P(R)=7/20, P(G)=5/20, P(B)=8/20. Hence the desired probability is, P = (7/20)(5/20)(8/20) = 7(5)(8) / 20^3 = 280/8000 = 7/200 = 0.035. Conclusion: P(R,G,B) = 0.035.

#2) Here we still need permutation (7,5,8), but s/replacement, thus reducing the pot by one marble with each draw of the hat. So, we have:

Again a permutation on the same set but s/replacement,
P(R,G,B) = (7/20)(5/19)(8/18) = 7/(19*9) = 7/171 = 0.041 to three places.

#3) I assume three marbles drawn again but, to answer as worded, P(R>1)=1. This is not quite true as you do not stipulate number. The "randomly select marbles" bit, painted, for me at least, marbles -- casually drawn until the cows done come home. Of course, had the cattle strolled with similar random disregard, I shutter to udder the time would abound, for even a hound would lie dead on the ground, done struck square in his head by them marbles of red.

If you don't like that, try [7c3 + 7c2*13] / (20c3) with a decimal approximation of 0.270.

Enjoy,

Rich B.
• May 28th 2006, 02:45 PM
Soroban
Hello, ptiva!

Quote:

A bag contains 7 red marbles, 5 green marbles, and 8 blue marbles.

a) If you randomly select 3 marbles with replacement,
what is the probability of drawing a red, a green and a blue marble in that order?
Since the drawing is with replacement, the bag contains all the marbles at each draw.

$P(1st\,R) = \frac{7}{20}$
$P(2nd\,G) = \frac{5}{20}$
$P(3rd\,B) = \frac{8}{20}$

Therefore: $P(RGB) \:= \:\frac{7}{20} \times \frac{5}{20} \times \frac{8}{20} \:=\:\frac{280}{8000} \:= \:0.035
$

Quote:

b) If you randomly select 3 marbles without replacement,
what is the probability of drawing a red, a green, and a blue marble in that order?
Without replacement, the number of marbles decreases with each draw.

$P(1st\,R) = \frac{7}{20}$
$P(2nd\,G) = \frac{5}{19}$
$P(3rd\,G) = \frac{8}{18}$

Therefore: $P(RGB) \ =\ \frac{7}{20} \times\frac{5}{19}\times\frac{8}{18}\ =\ \frac{280}{6840}\ \approx\ 0.041$

Quote:

c) If you randomly select 3 marbles without replacement,
what is the probability that at least 2 of them will be red?
"At least 2 Red" means: 2 Red or 3 Red.

Two Reds

One possible order is: Red-Red-Other

. . $P(RRO) \:= \:\frac{7}{20} \times \frac{13}{19}\times\frac{12}{18} \:= \:\frac{546}{6840}$

Since there are three possible orders: $\{RRO,\;ROR,\;ORR\}$

. . then $P(exactly\,2R) = 3\times\frac{546}{6840}\:=\:\frac{1638}{6840}$

Three Reds

$P(3R)\:=\:\frac{7}{20}\times\frac{6}{19}\times \frac{5}{18}\:=\:\frac{210}{6840}$

Therefore: $P(2R\,or\,3R) \:=\:\frac{1638}{6840} + \frac{210}{6840} \:= \:\frac{1848}{6840}\;\approx\;0.2702$