Question

3 animals need to be chosen. There are 3 tigers, 6 bears, and 4 dogs. Find the probability that:

a) 1 tiger, 1 bear, and 1 dog are chosen
b) 3 bears are chosen
c) no tigers are chosen

by the way, we are learning combinations and permutations.

Thank you for your help!!!

Answer 1

You have probably learned
C(n,r)=n!/((n-r)!r!)
is the number of combinations of choosing r objects out of n.

So in general, if we have letters
a,a,a, b,b, c,c,c,c
there are
C(3,1) ways to choose a letter a
C(2,1) ways to choose a letter b and
C(4,1) ways to choose a letter c
as opposed to
C(9,3) ways to choose any three letters.
So the probability of choosing one of each letter is
C(3,1)*C(2,1)*C(4,1)/C(9,3)
The formula is easy to remember because
3+2+4=9, 1+1+1=3 in the respective positions.

Parts (a) and (b) can be similarly solved with straight application of the formula above, substituting the appropriate letters:
t,t,t, b,b,b,b,b,b, d,d,d

For part (c), bears and dogs can be treated as one group (not tigers) of 10 animals, so the equation would be reduced to
P(no tigers)=C(10,3)*C(3,0)/C(13,3)

Answer 2

probability = (number of favourable outcomes)/(total number of outcomes)

total outcomes = 3+6+4 = 13
Question 1:

probability tiger = 3/13
then
probability bear = 6/12
(as one from the 13 has been picked)
then
probability dog = 4/11
(as another from the remaining 12 has been picked).

Prob(tiger and bear and dog) = multiply the probabilities together

3/13 * 6/12 * 4/11 =

Three bears

P(first bear) = 6/13
P(second bear) = 5/12
P(third bear) = 4/11

Multiply all three to get
P(three bears)

Probability(no tigers) =

10/13 * 9/12 * 8/11

10 is the bears + dogs