1. ## Please check my probability work guys...

Hey guys, I'm having some problems with combinations and permutations.

I have included my answers as well, explanations will be highly appreciated.

I'll try to give +rep+ if it is possible.

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1. The first three digits of a telephone exchange are 452.

If all the sequences of the remaining digits are equally likely, what is the probability that a randomly selected phone number contains seven distinct digits?

My answer: $\frac{7 \cdot 6 \cdot 5 \cdot 4}{10!} = \frac{1}{4320}$

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2. How many different meals can be made from four kinds of meat, six vegetables, and three starches if a meal consists of one selection from each group?
My answer: $\frac{13!}{4!6!3!} = 60 060$

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3. A woman getting dressed for a night out is asked by her significant other to wear a dress, high heeled sneakers, and a wig. In how many orders can she put on these objects?

My answer: $3! = 6$

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4. Two dice are rolled, and the sum of the face values is six. What is the probability that at least one of the dice came up a 3?

My answer: $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$

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5. Answer problem 4 again, given that the sum is less than six.

My answer: Erm... Not sure. Maybe: $1 - 0,333 = \frac{2}{3}$ ?

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6. A couple has two children. What is the probability that both are girls given that the oldest is a girl? What is the probability that both are girls given that one of them is a girl?

My answer: Isn't that the same thing? One is always a girl?

$\frac{1!}{2!} = 0,5$

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7. A box has three coins. One has two heads, one has two tails, and the last is a fair coin with one head and one tail. A coin is chosen at random, is flipped, and comes up heads.

a) What is the probability that the chosen coin is the two headed coin?

b) What is the probability that if it is thrown another time it will come up heads?

c) Answer part a) again, supposing that the coin is thrown a second time and comes up heads again.

a) $\frac{1!}{2!} = 0,5$ (Not sure about this one though...)

b) If it's the fair coin, 50%
....If it's the two headed one, 100%

But how do we tell???

c) Erm... Still 0,5?

2. Hello, janvdl!

1) The first three digits of a telephone exchange are 452.

If all the sequences of the remaining digits are equally likely, what is the probability
that a randomly selected phone number contains seven distinct digits?
There are 10,000 possible four-digit numbers (from 0000 to 9999).

Since 4, 5, and 2 are already used, there are seven digits available.
. . The first digit can be any of the 7 available digits.
. . The second digit can be any of the 6 remaining digits.
. . The third digit can be any of the 5 remaining digits.
. . The fourth digit can be any of the 4 remaining digits.
Hence, there are: . $7\cdot6\cdot5\cdot4 \:=\:840$ numbers with seven distinct digits.

Therefore: . $P(\text{7 distinct digits}) \;=\;\frac{840}{10,000} \;=\;\frac{21}{250}$

2. How many different meals can be made from 4 kinds of meat, 6 vegetables,
and 3 if a meal consists of one selection from each group?
Number of meals is: . $4 \times 6 \times 3 \:=\:72$

3. A woman getting dressed for a night out is asked by her significant other
to wear a dress, high heeled sneakers, and a wig.
In how many orders can she put on these objects?

My answer: $3! = 6$ . . . . Right!

4. Two dice are rolled, and the sum of the face values is six.
What is the probability that at least one of the dice came up a 3?
We are told that sum is six.
There were five possible outcomes: . $(1,5),\2,4),\3,3),\4,2),\5,1)" alt="(1,5),\2,4),\3,3),\4,2),\5,1)" />

Among them only one has a "3".

Therefore: . $Prob. \:=\:\frac{1}{5}$

5. Answer problem 4 again, given that the sum is less than six.
We are told that the sum is less than six.

There are eleven possible outcomes:
. . $(1,1),(1,2),(1,3),(1,4), (2,1), (2,2), (2,3), (3,1), (3,2), (1,4),(4,1)$

Among them, four of them contain a "3".

$Prob. \:=\:\frac{4}{11}$

6. A couple has two children.
(a) What is the probability that both are girls given that the oldest is a girl?
(b) What is the probability that both are girls given that one of them is a girl?

(a) The oldest child is a girl.

The probability that the younger child is also a girl is $\frac{1}{2}$

That is: . $P(\text{both girls }|\text{ older is a girl}) \;=\;\frac{1}{2}$

(b) This is a classic trick question.

We are told that one of the children is a girl.
Then there are three possible cases . . .

[1] The older is a boy, the younger is a girl.
[2] The older is a girl, the younger is a boy.
[3] Both are girls.

Among the three cases, only one has two girls.

$P(\text{both girls }|\text{ one is a girl}) \;=\;\frac{1}{3}$

3. Originally Posted by Soroban
This is a classic trick question.

We are told that one of the children is a girl.
Then there are three possible cases . . .

[1] The older is a boy, the younger is a girl.
[2] The older is a girl, the younger is a boy.
[3] Both are girls.

Among the three cases, only one has two girls.

$P(\text{both girls }|\text{ one is a girl}) \;=\;\frac{1}{3}$

Man, that one really is a trick question!

Soroban, thanks a lot for the answers , I'll go through it all again now.
I'm sure I'll get the hang of these things soon.

Thanks again, really appreciate it.

4. 1. A telephone number is 7 digits long and the first three are given as 4,2,5. So the possibilities are (452)-xxxx. There are 7 choices for for the first x if you allow for a 0 in that position (0,1,3,6,7,8,9), 6 for the second, 5 for the third, 4 for the 4th. So I agree that there are 7*6*5*4 possible phone numbers starting with 452 that contain each digit only once. However the total number of 7 digit numbers is 10^7 (if we allow for any digit in any position). If we are restricting ourselves to the those numbers with 452 as the first three digits there are only 10^4 total such phone numbers.

2. 4*6*3 = 72

3. Correct

4. How many ways are there for two dice to sum to 6??
(dice 1, dice 2): (1,5), (2,4), (3,3), (4,2), (5,1)
1/5 rolls contain a 3.

5. possibilities are:
(1,1)
(1,2) (2,1)
(1,3) (3,1) (2,2)
(1,4) (4,1) (3,2) (2,3)
4 contain at least one 3, so 4/10 = 2/5.

6. Let A be the event that first born is a girl . B is the event that second born is a girl. Prob(A and B) is the probability that both first and second born are girls. Prob(A | B) is the probability of A given B and is given by the conditional probability formula Prob(A | B) = Prob(A and B)/Prob(B).

We are first asked to find the probability that A and B happened given that A happened:
Prob(A and B | A) = Prob(A and B and A)/Prob(A) = Prob(A and B)/Prob(A) =Prob(A)*Prob(B)/Prob(A) = Prob(B) = 1/2

We are asked then asked to find the probability that A and B happen given either A happened or B happened:
P(A and B | A or B)
= Prob((A and B) and (A or B))/Prob(A or B)
= Prob(A and B)/Prob(A or B)
= Prob(A)Prob(B)/(Prob(A)+Prob(B)-Prob(A and B))
= (1/2 * 1/2)/(3/4)
= 1/3

We can see why this is: Consider - (girl,girl) (girl,boy) (boy,girl) (boy,boy) as all possible outcomes. If we know that at least one of them is a girl all we care about are: (girl, girl) (girl, boy) and (boy, girl). (girl girl) happens once in three possibilities so the probability is 1/3.

5. Originally Posted by iknowone
2. Correct
Thank you! However you and Soroban seem to disagree over number 2...

6. ## correction.

I apologize, I misread the question as allowing meals to be composed of 3 of any of the above, without the restriction of one from each. Soroban is correct since you can choose 1 of 4 different meats and 1 of 6 different vegetables, etc.