# Math Help - Further Probability problems

1. ## Further Probability problems

1)A bag contains five white, six red and seven blue balls. If three balls are selected at random, without replacement, find the probability they are:

-All different colors

2)A four 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:
a)is even
b)is odd
c)is less than 4000
d)is less than 4000, given that it is greater than 3000

2. 1) $\displaystyle 18$ balls in total. So the probability of getting all different colours would be:

$\displaystyle \frac{5}{18} \times \frac{6}{17} \times \frac{7}{16} = \frac{35}{816}$.

2)
a) You require the last digit to be even. So the probability is $\displaystyle \frac{4}{8} = \frac{1}{2}$.

b) You require the last digit to be odd. So the probability is $\displaystyle \frac{4}{8} = \frac{1}{2}$.

c) You require the first digit to be a 0, 1, 2, 3. So the probability is $\displaystyle \frac{4}{8} = \frac{1}{2}$.

d) There are $\displaystyle {8\choose{4}} = 70$ different combinations of 4 numbers from the 8 given.

It has to be greater than 3000 and less than 4000. This means you know the first digit has to be a 3 and there are 7 digits left to choose from.

That means you will be finding the number of combinations of 3 digits from the 7 remaining.

There are $\displaystyle {7\choose{3}} = 35$ combinations.

That means the probability of having a number less than 4000 and greater than 3000 is $\displaystyle \frac{35}{70} = \frac{1}{2}$.

3. Hello, johnsy123!

$\text{1)A bag contains five white, six red and seven blue balls.}$
$\text{If three balls are selected at random, without replacement,}$
$\text{find the probability they are all different colors.}$

There are: . ${18\choose3} \:=\:816$ possible outcomes.

There are: . ${5\choose1}{6\choose1}{7\choose1} \:=\:5\cdot6\cdot7 \:=\:210$ ways to get one of each color.

Therefore: . $P(\text{all different colors}) \;=\;\dfrac{210}{816} \;=\;\dfrac{35}{136}$

$\text{2) A four-digit number (no repetitions) is to be formed}$
$\text{from the set of numbers (0,1,2,3,4,5,6,7).}$

Assuming that the number may not have a leading zero,
. . there are: . $7\cdot7\cdot6\cdot5 \:=\:1470$ possible 4-digit numbers.

$\text{(a) Find the probability that the number is even.}$

There are two cases to consider:
. . The last digit is 0.
. . The last digit is 2, 4 or 6.

The last digit is 0: . $\square\:\square\;\square\;0$
. . The other 3 spaces can be filled in: . $7\cdot6\cdot5 \:=\:210$ ways.

The last digit is 2, 4 or 6: .3 choices.
The first must not be 0: .6 choices.
The other 2 spaces can be filled in $6\cdot5\,=\,30$ ways.
. . There are: . $3\cdot6\cdot30 \:=\:540$ ways.

Hence, there are: . $210 + 540 \:=\:750$ even numbers.

Therefore: . $P(n\text{ is even}) \:=\:\dfrac{750}{1470} \;=\;\dfrac{25}{49}$

$\text{(b) }n\text{ is odd.}$

$P(n\text{ is odd}) \;=\;1 - \dfrac{25}{49} \;=\;\dfrac{24}{49}$

$\text{(c) }n\text{ is less than 4000.}$

The first digit must be 1, 2 or 3: .3 choices.
The other three digits may be placed in $7\cdot6\cdot5 \:=\:210$ ways.
. . There are: . $3\cdot210 \:=\:630$ numbers less than 4000.

Therefore: . $P(n < 4000) \;=\;\dfrac{630}{1470} \;=\;\dfrac{3}{7}$

$\text{(d) }n\text{ is less than 4000, given that it is greater than 3000}$

We want: . $P(n < 4000\,|\,n > 3000)$

$\text{Bayes' Theorem: }\;P(n < 4000\,|\,n > 3000) \;=\;\dfrac{P([n < 4000]\, \wedge\, [n > 3000])}{P(n > 3000)}$ .[1]

The numerator is: . $P(3000 < n < 4000)$

The first number must be 3.
The other three digits can be placed in $7\cdot6\cdot5 \:=\:210$ ways.
There are $210$ numbers between 3000 and 4000.
. . $P(3000 < n < 4000) \:=\:\dfrac{210}{1470} \:=\:\dfrac{1}{7}$

The denominator is: . $P(n > 3000)$

The first digit is 3, 4, 5, 6, or 7: .5 choices.
The other three digits can be placed in $7\cdot6\cdot5 \:=\:210$ ways.
There are $5\cdot210 \:=\:1050$ numbers greater than 3000.
. . $P(n > 3000) \:=\:\dfrac{1040}{1470} \;=\;\dfrac{5}{7}$

Substitute into [1]:

. . $P(n < 4000\,|\,n > 3000) \;=\;\dfrac{\frac{1}{7}}{\frac{5}{7}} \;=\;\dfrac{1}{5}$

4. This is why I HATE probability with a passion :P

Thanks Soroban.