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!!!
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)
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
To find the probabilities, we first need to find the total number of possible outcomes, which is the total number of ways to choose 3 animals from the given options.
a) To choose 1 tiger, 1 bear, and 1 dog, we need to consider the individual choices for each animal and then multiply them together.
Number of ways to choose 1 tiger = 3
Number of ways to choose 1 bear = 6
Number of ways to choose 1 dog = 4
Total number of ways = 3 x 6 x 4 = 72
So, there are 72 possible outcomes when choosing 1 tiger, 1 bear, and 1 dog.
b) To choose 3 bears, we only have 6 bears to choose from. Since we need to choose 3 of them, we use the combination formula:
Number of ways to choose 3 bears = C(6, 3) = 6! / (3! * (6-3)!) = 6! / (3! * 3!) = (6 * 5 * 4) / (3 * 2 * 1) = 20
So, there are 20 possible outcomes when choosing 3 bears.
c) To choose no tigers, we can only choose from the bears and dogs. So, the number of ways to choose 3 animals without any tiger = number of ways to choose 3 animals from 6 bears and 4 dogs.
Number of ways to choose 3 animals = C(10, 3) = 10! / (3! * (10-3)!) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120
So, there are 120 possible outcomes when choosing no tigers.
Finally, to find the probabilities, we divide the desired outcomes by the total possible outcomes.
a) Probability of choosing 1 tiger, 1 bear, and 1 dog = 72 / (72 + 20 + 120) = 72 / 212 ≈ 0.3396
b) Probability of choosing 3 bears = 20 / 212 ≈ 0.0943
c) Probability of choosing no tigers = 120 / 212 ≈ 0.5660
To find the probabilities, we first need to determine the total number of possible outcomes and the number of favorable outcomes for each scenario.
Total Possible Outcomes:
The total number of possible outcomes can be found by adding up the total number of tigers, bears, and dogs. In this case, there are 3 tigers + 6 bears + 4 dogs = 13 animals in total.
Favorable Outcomes:
a) To find the number of favorable outcomes for scenario a (1 tiger, 1 bear, and 1 dog), we need to determine the number of ways to choose 1 tiger, 1 bear, and 1 dog from their respective groups.
The number of ways to choose 1 tiger from 3 is given by the combination formula, which is denoted as C(3, 1) = 3.
Similarly, the number of ways to choose 1 bear from 6 is C(6, 1) = 6, and the number of ways to choose 1 dog from 4 is C(4, 1) = 4.
Since we want to find the probability of all three events happening together, we use the multiplication principle. Therefore, the favorable outcomes for scenario a is calculated as 3 * 6 * 4 = 72.
b) For scenario b (3 bears are chosen), we need to determine the number of ways to choose 3 bears from the group of 6 bears. This can be calculated as C(6, 3) = 20.
c) To determine the number of favorable outcomes for scenario c (no tigers are chosen), we need to find the number of ways to choose 3 animals (either bears or dogs) from their respective groups without choosing any tigers.
The number of ways to choose 3 bears from 6 can be calculated as C(6, 3) = 20.
The number of ways to choose 3 dogs from 4 can be found using C(4, 3) = 4.
Since we only want the favorable outcomes where no tigers are chosen, we need to add the favorable outcomes for choosing 3 bears and choosing 3 dogs together. So the total number of favorable outcomes for scenario c is 20 + 4 = 24.
Calculating Probabilities:
To find the probability of an event, we divide the number of favorable outcomes by the total number of possible outcomes.
a) Probability of 1 tiger, 1 bear, and 1 dog being chosen: P(a) = favorable outcomes / total outcomes = 72 / 13 = 5.54 (rounded to two decimal places).
b) Probability of 3 bears being chosen: P(b) = favorable outcomes / total outcomes = 20 / 13 = 1.54 (rounded to two decimal places).
c) Probability of no tigers being chosen: P(c) = favorable outcomes / total outcomes = 24 / 13 = 1.85 (rounded to two decimal places).
Therefore, the probabilities for each scenario are:
a) The probability of choosing 1 tiger, 1 bear, and 1 dog is approximately 5.54%.
b) The probability of choosing 3 bears is approximately 1.54%.
c) The probability of not choosing any tigers is approximately 1.85%.