Probability with letter permutations!

• Mar 21st 2010, 10:05 AM
ibetan
Probability with letter permutations!
13. Suppose a bag contains the letters to spell
probability.

a) How many four-letter arrangements are
possible using these letters?
I made 4 separate cases (no double letters, i doubles, b doubles, both b and i doubles) and added them to get 3074 ways. Dunno if this is correct
b) What is the probability that Barb
chooses four letters from the bag in the
order that spell her name?
2/3074

d) What four-letter arrangement would be
most likely to be picked? Explain your
reasoning.
One containing at least one i and b.

Can anyone tell me if i did this correctly, the solutions does not have the answer for this question.
• Mar 21st 2010, 11:07 AM
awkward
Quote:

Originally Posted by ibetan
13. Suppose a bag contains the letters to spell
probability.

a) How many four-letter arrangements are
possible using these letters?
I made 4 separate cases (no double letters, i doubles, b doubles, both b and i doubles) and added them to get 3074 ways. Dunno if this is correct
b) What is the probability that Barb
chooses four letters from the bag in the
order that spell her name?
2/3074

d) What four-letter arrangement would be
most likely to be picked? Explain your
reasoning.
One containing at least one i and b.

Can anyone tell me if i did this correctly, the solutions does not have the answer for this question.

Ibetan,

I get a different answer on a)-- so I suggest you post the details of your solution and then we can see where we differ.
• Mar 21st 2010, 11:25 AM
Soroban
Hello, ibetan!

I assume "arrangements" means that the order of the letters is considered.
That is, we are spelling four-letter "words".

Quote:

13. Suppose a bag contains the letters to spell PROBABILITY.

a) How many four-letter arrangements are possible using these letters?

There are four cases:
. . (1) BB and II
. . (2) BB only
. . (3) II only
. . (4) No doubles

(1) BB and II

There are: . ${4\choose2,2} \,=\,6$ arrangements with $\{B,B,I,I\}$

(2) BB only

We have: . $\{B,B,X,Y\}$
Choose two different letters from the other 8 letters: . ${8\choose2} = 28$ ways.
Then $\{B,B,x,y\}$ can be arranged in: . $\frac{4!}{2!} \,=\,12$ ways.

Hence, there are: . $28\cdot12 \,=\,336$ arrangements with $BB$ only.

(3) II only

We can use the same reasoning from Case (2).

Hence, there are: . $336$ arrangements with $II$ only.

(4) No doubles

Select and arrange four letters from the nine different letters.

Hence, there are: . $_9P_4 \,=\,3024$ arrangements with four different letters.

Therefore, there are: . $6 + 336 + 336 + 3924 \:=\:3702$ possible arrangements.

Quote:

b) What is the probability that Barb chooses four letters from the bag
in the order that spell her name?

She chooses $B,A,R,B$ in that order.

. . $\begin{array}{ccc}P(B) &=& \frac{2}{11} \\ \\[-3mm]
P(A) &=& \frac{1}{10} \\ \\[-3mm]
P(R) &=& \frac{1}{9} \\ \\[-3mm]
P(B) &=& \frac{1}{8} \end{array}$

Therefore: . $P(B,A,R,B) \;=\;\frac{2}{11}\cdot\frac{1}{10}\cdot\frac{1}{9} \cdot\frac{1}{8} \;=\;\frac{1}{3960}$

Quote:

d) What four-letter arrangement would be most likely to be picked?
A strange question . . . It could have been worded more precisely.

There are 3702 possible four-letter arrangements.
. . And 3024 of them has four different letters.

So, selecting four letters at random,
. . you would get four different letters over 80% of the time.

Which set of four different letters is the most likely?
. . Um . . . all of them?

• Mar 21st 2010, 06:20 PM
ibetan
Quote:

Originally Posted by awkward
Ibetan,

I get a different answer on a)-- so I suggest you post the details of your solution and then we can see where we differ.

Both my cases for one set of double letters were wrong.(Headbang)