# Math Help - 3 Questions regarding probability (Counting Methods)

1. ## 3 Questions regarding probability (Counting Methods)

First Question:
A 4 digit number ( with no repetitions) is to be formed from the set of numbers {0,1,2,3,4,5,6,7}. Find the probability that the number is even.

_ _ _ _ <----There are 4 places u have to fill with numbers. last spot can occupy 4P1 ways. 1st spot can occupy 6P1 spots, (since 0 cant be on the 1st spot) and then the 2 spots left can be occupied in 6P2 ways.

However there can also be another set of possibility. the last spot can be filled 4P1 ways, but lets say 0 was picked for the last spot hence there are 7P1 ways of occupying the 1st spot and 6P2 for the last 2.

This gives: 4P1 x 6P1 x 6P2 + 4P1 x 7P1 x 6P2 = 1560 ways

then the total ways of occupying the 4 digits with no restrictions is 7P1 x 7P3 = 1470. 7P1 because for the 1st spot because it cant be 0. and the rest of the 3 spots have 7P3 ways of occupying.

however... there must be a mistake in my working since i got more ways of making a even number than the total number of ways with no restrictions lol.... can some1 just explain through how they would do it? Thanks

Second Question:
A hand of 5 cards is dealt from a normal pack of 52 cards. Find the probability that the hand will contain the ace of spades given that there is at least one ace.

so Pr(AofS | more than 1A)= Pr(AofS n more than 1A) / Pr (more than 1A)
but how do u work out Pr(AofS n more than 1A)? ive tried some ways but they didnt quite work. Could some1 please explain through each step how they would do this question. Thanks

Third Question:
Of the intergers from 1000 to 9999 how many have at least one digit a 5 or 7?
not too sure how to approach this. i did do some workings but dont think it will help if i posted it up here. So could some1 just show their working on how they would do it. Thanks!

Many thanks to anyone who can help!

2. Hello,
Originally Posted by TrueTears
First Question:
A 4 digit number ( with no repetitions) is to be formed from the set of numbers {0,1,2,3,4,5,6,7}. Find the probability that the number is even.

_ _ _ _ <----There are 4 places u have to fill with numbers. last spot can occupy 4P1 ways. 1st spot can occupy 6P1 spots, (since 0 cant be on the 1st spot) and then the 2 spots left can be occupied in 6P2 ways.

However there can also be another set of possibility. the last spot can be filled 4P1 ways, but lets say 0 was picked for the last spot hence there are 7P1 ways of occupying the 1st spot and 6P2 for the last 2.

This gives: 4P1 x 6P1 x 6P2 + 4P1 x 7P1 x 6P2 = 1560 ways

then the total ways of occupying the 4 digits with no restrictions is 7P1 x 7P3 = 1470. 7P1 because for the 1st spot because it cant be 0. and the rest of the 3 spots have 7P3 ways of occupying.

however... there must be a mistake in my working since i got more ways of making a even number than the total number of ways with no restrictions lol.... can some1 just explain through how they would do it? Thanks
For the number to be even, it's just sufficient that the last digit is 0,2,4 or 6. That makes $4$ possibilities.
Say you choose one of them. There remain 7 possible digits for the last 3 numbers.
Let's take care of the 3rd. There are 7 possible.
Let's take care of the 2nd. There are 6 possible.
Let's take care of the 1st. There are 5 possible.
The numbers of possibilities is hence $4 \times 7 \times 6 \times 5$

The overall number of possibilities is $8 \times 7 \times 6 \times 5$ (no repetition, so each time you pick a number, you can't pick it again).

So the probability is $\frac{4 \times 7 \times 6 \times 5}{8 \times 7 \times 6 \times 5}=\frac 12$

Actually, it could have been more straightforward, because there are as many even numbers as odd numbers that are 4-digit without repetition !

I'll try the other ones later

3. ah i see wat you're saying but the book's answer says its 25/49 o.o lol

4. Originally Posted by TrueTears
ah i see wat you're saying but the book's answer says its 25/49 o.o lol
Even 4 digit number:

With zero at the end the number of arrangements is (7)(6)(5)(1) = 210.

Without zero at the end the number of arrangements is (6)(6)(5)(3) = 540.

Note: 3 choices for the last digit, 6 choices for the first digit (6 non-zero numbers to choose the first digit from once the last digit is chosen, 6 numbers [including zero] to choose the second digit from once the last and first digits are chosen, 5 numbers to choose the third digit from once the last, first and second digits are chosen).

Total four digit even numbers = 750.

Number of four digit numbers without restriction of being even is (7)(7)(6)(5) = 1470.

Pr(even number) = 750/1470 = ......

5. Originally Posted by mr fantastic
Even 4 digit number:

With zero at the end the number of arrangements is (7)(6)(5)(1) = 210.

Without zero at the end the number of arrangements is (6)(6)(5)(3) = 540.

Note: 3 choices for the last digit, 6 choices for the first digit (6 non-zero numbers to choose the first digit from once the last digit is chosen, 6 numbers [including zero] to choose the second digit from once the last and first digits are chosen, 5 numbers to choose the third digit from once the last, first and second digits are chosen).

Total four digit even numbers = 750.

Number of four digit numbers without restriction of being even is (7)(7)(6)(5) = 1470.

Pr(even number) = 750/1470 = ......
Oh right, I forgot the case when the first digit is 0, that wouldn't make it a 4 digit number...
^^
Sorry to the OP for the confusion !

Second Question:
A hand of 5 cards is dealt from a normal pack of 52 cards. Find the probability that the hand will contain the ace of spades given that there is at least one ace.

so Pr(AofS | more than 1A)= Pr(AofS n more than 1A) / Pr (more than 1A)
but how do u work out Pr(AofS n more than 1A)? ive tried some ways but they didnt quite work. Could some1 please explain through each step how they would do this question. Thanks
Okay, let's do this one.

Let's define the events :
B = There is at least 1 ace in your hand. Note that its complement is "there is no ace in your hand"
Let E be the probability space (that is $P(E)=1$ and $A \cap E=A \quad \forall A$)

We have a general formula : $\boxed{P(A)=P(A \cap B)+P(A \cap B^c)}$
$\text{Proof : } A=A \cap E=A \cap (B \cup B^c)=(A \cap B) \cup (A \cap B^c)$

$\implies P(A)=P(A \cap B)+P(A \cap B^c)-P\underbrace{(A \cap B \cap A \cap B^c)}_{\emptyset \text{ because } B \cap B^c=\emptyset}\quad \blacksquare$

But $A$ and $B^c$ cannot occur at the same time, because it means that "there is no ace in your hand" and "there is the ace of spades in your hand", which is not possible.
Hence $P(A \cap B^c)=0$

So $\boxed{P(A)=P(A \cap B)}$

------------------------------------------------
Therefore
$P(A/B)=\frac{P(A \cap B)}{P(B)}=\frac{P(A)}{P(B)}$

What is $P(A)$, the probability that you have an ace of spades in your hand ?
The number of hands containing 5 random cards is $C_{52}^5$

The number of hands containing the ace of spades (event A) is : $C_{51}^4$ (suppose you take the ace of spades. There remain 4 cards to pick among 51)
So $P(A)=\frac{C_{51}^4}{C_{52}^5}$

The number of hands containing no ace (event $B^c$) is :
$C_{48}^5$ (suppose you remove all the aces since you don't want them. There remain 5 cards to take among 48)
So $P(B^c)=\frac{C_{48}^5}{C_{52}^5} \implies P(B)=1-\frac{C_{48}^5}{C_{52}^5}$

Is it clear ?

Side note : $C_n^k=\frac{n!}{k! (n-k)!}={n \choose k}$