Explanation:
a probability is always the ratio
desired cases / totally possible cases
10 black balls
7 red balls
8 white balls
in total 25 balls.
a)
we have the possible outcomes
w r b
w b r
r w b
r b w
b w r
b r w
now we need to identify the individual probabilities of each draw, multiply the fitting 3 draws per outcome, and add these 6 numbers (as they are independent outcomes without any overlap).
w r b
P(white ball on first draw) = 8/25
P(red ball on second draw) = 7/24 (as there are now only 24 balls available for the second draw)
P(black ball on the third draw) = 10/23 (only 23 balls left for the third draw)
P (w r b) = 8/25 × 7/24 × 10/23 = 1/5 × 7/3 × 2/23 =
= 14/345 = 0.04057971...
w b r
P(w 1st) = 8/25
P(b 2nd) = 10/24
P(r 3rd) = 7/23
P(w b r) = 8/25 × 10/24 × 7/23 = 14/345 = 0.04057971...
r w b
P(r 1st) = 7/25
P(w 2nd) = 8/24
P(b 3rd) = 10/23
P(w b r) = 7/25 × 8/24 × 10/23 = 14/345 = 0.04057971...
you see what is going on right ? we are simply mixing the same 3 numbers in the numerator and the other same 3 numbers in the denominator, so that the product of the 3 fractions is always the same.
so, all 6 outcome probabilities are the same.
therefore, the probability that exactly one ball is white, and exactly one ball is red, and exactly one ball is black is
6 × 0.04057971... = 6×14/345 = 2×14/115 = 28/115 =
= 0.243478261...
b)
in a similar way we have the desired outcomes
b b w
b b r
b w b
b r b
w b b
r b b
b b w
P(b 1st) = 10/25
P(b 2nd) = 9/24 (remember, without replacement, so there are only 6 of total 24 balls black)
P(w 3rd) = 8/23
P(b b w) = 10/25 × 9/24 × 8/23 = 2/5 × 3 × 1/23 =
= 6/115 = 0.052173913...
b b r
P(b 1st) = 10/25
P(b 2nd) = 9/24
P(r 3rd) = 7/23
P(b b r) = 10/25 × 9/24 × 7/23 = 1/5 × 3/4 × 7/23 =
= 21/460 = 0.045652174...
b w b
P(b w b) = P(b b w) = 6/115 = 0.052173913... (for the same train as why the probabilities under a) were equal).
b r b
P(b r b) = P(b b r) = 21/460 = 0.045652174...
w b b
P(w b b) = P(b b w) = 6/115 = 0.052173913...
r b b
P(r b b) = P(b b r) = 21/460 = 0.045652174...
the probability that there are exactly 2 black balls is
3×6/115 + 3×21/460 = 18/115 + 63/460 =
= 0.293478261...
c)
this is the complementary probability that there is at least one red ball.
P(no red) = 1 - P(at least 1 red)
our outcomes for at least 1 red
r w w
r b b
r w b
r b w
w r w
b r b
w r b
b r w
w w r
b b r
w b r
b w r
r r w
r r b
r w r
r b r
w r r
b r r
r r r
the equal groups (with the same 3 numerator and the save 3 denominators) are
r w w, w r w, w w r
r b b, b r b, b b r
r w b, r b w, b r w, b w r, w r b, w b r
r r w, r w r, w r r
r r b, r b r, b r r
r r r
P(2w 1r) = 8/25 × 7/24 × 7/23 = 1/25 × 7/3 × 7/23 =
= 49/1725 = 0.028405797...
P(2b 1r) = 10/25 × 9/24 × 7/23 = 1/5 × 3/4 × 7/23 =
= 21/460 = 0.045652174...
P(1r 1w 1b) = 7/25 × 8/24 × 10/23 = 14/345 (see a)) =
= 0.04057971...
P(2r 1w) = 7/25 × 6/24 × 8/23 = 7/25 × 1 × 2/23 =
= 14/575 = 0.024347826...
P(2r 1b) = 7/25 × 6/24 × 10/23 = 7/5 × 1/2 × 1/23 =
= 7/230 = 0.030434783...
P(3r) = 7/25 × 6/24 × 5/23 = 7/5 × 1/4 × 1/23 =
= 7/460 = 0.015217391...
the total probability of having at least one red ball is then
3×P(2w 1r) + 3×P(2b 1r) + 6×P(1w 1r 1b) + 3×P(2r 1w) + 3×P(2r 1b) + P(3r)
the probabilty of having no red ball at all is then
1 - (3×P(2w 1r) + 3×P(2b 1r) + 6×P(1w 1r 1b) + 3×P(2r 1w) + 3×P(2r 1b) + P(3r)) =
1 - 3×49/1725 - 3×21/460 - 6×14/345 - 3×14/575 - 3×7/230 - 7/460 = 0.354782609...
d)
we have only one outcome
r r w
the probabilty to have first 2 red balls and then 1 white ball is
P(r r w) = 7/25 × 6/24 × 8/23 = 7/25 × 2 × 1/23 =
= 14/575 = 0.024347826...