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 June 12th, 2011, 07:39 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability 1.Three letters are selected at random from the word BIOLOGY.Find the number of ways in doing this.Hence,find the probability that the selection a)Does not contain letter O b)Contains both of the letter O Three letter select at random from the word BIOLOGY, i using 7C3 But the solution provided is 5C3 + 5C2 + 5C1 =25 I don understand why it is like that,and part (a) and (b) too. 2.Four person are chosen at random from a group of ten person consisting of four men and six women.Three of the women are sisters.Calculate the probability that the four person chosen will a)include the three sisters. This is the last part of the question,the other part i am able to solve.But having problem in this part.Thankk you.
June 12th, 2011, 08:02 AM   #2
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Re: Probability

Hello, hoyy1kolko!

Quote:
 2. Four person are chosen at random from a group of ten person consisting of four men and six women. Three of the women are sisters. Calculate the probability that the four person chosen will (a) include the three sisters.

Four people are chosen from the available ten people.
[color=beige]. . [/color]$\text{There are: }\:_{10}C_4 \:=\:210\text{ possible choices.}$

There are four men:[color=beige] .[/color]$\{M_1,\,M_2,\,M_3,\,M_4\}$
And six women, three of whom are sisters:[color=beige] .[/color]$\{S_1,\,S_2,\,S_3,\,W_1,\,W_2,\,W_3\}$

There is 1 way to choose the 3 sisters.
$\text{And }\,_7C_1 \,=\,7\text{ choices for the fourth person.}$

$\text{Therefore: }\:P(\text{3 sisters}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$

June 12th, 2011, 08:40 AM   #3
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Re: Probability

Quote:
 Originally Posted by hoyy1kolko 1.Three letters are selected at random from the word BIOLOGY.Find the number of ways in doing this.Hence,find the probability that the selection a)Does not contain letter O b)Contains both of the letter O Three letter select at random from the word BIOLOGY, i using 7C3 But the solution provided is 5C3 + 5C2 + 5C1 =25 I don understand why it is like that,and part (a) and (b) too.
The 3 letters you choose will contain (no O's or 1 O or 2 O's). That is where 5C3 + 5C2 + 5C1 comes from.

Perhaps a more general approach would be to use generating functions. For each of the letters B,I,L,G,Y either you don't choose it or you do choose it. Represent that as

$(1x^0+1x^1)(1x^0+1x^1)(1x^0+1x^1)(1x^0+1x^1)(1x^0+ 1x^1)=(1+x)^5$

For the O's you can either not choose any, choose 1, or choose 2. Represent this by

$(1x^0+1x^1+1x^2)=(1+x+x^2)$

Since you must do both tasks, multiply.

$(1+x)^5(1+x+x^2)=x^7+6x^6+16x^5+25x^4+25x^3+16x^2+ 6x+1$

and now you see every way to choose letters out of the seven. There are 25 ways to choose 3 letters, 16 ways to choose 5 letters, etc.

June 12th, 2011, 02:39 PM   #4
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Re: Probability

Hello, hoyy1kolko!

Quote:
 1. Three letters are selected at random from the word BIOLOGY. Find the number of ways in doing this. Three-letter select at random fromthe word BIOLOGY, I used 7C3. But the solution provided is: 5C3 + 5C2 + 5C1 = 25 I don't understand why.

mrtwhs gave a good explanation for their answer.

There are two O's and five Others.

No O's: Choose 3 letters from the 5 Others.
[color=beige]. . [/color]$_5C_3 \,=\,10\text{ ways to have no O#39;s.}$

One O: Choose one O and 2 Others.
[color=beige]. . [/color]$_2C_1 \,=\,2\text{ ways to choose one O}$
[color=beige]. . [/color]$_5C_2 \,=\,10\text{ ways to choose two Others}$
[color=beige]. . [/color]$\text{Hence: }\:2\,\times\,10 \:=\:20\text{ ways to have one O.}$

Two O's: Choose both O's and 1 Other.
[color=beige]. . [/color]$_2C_1\,=\,1\text{ way to choose both O#39;s.}$
[color=beige]. . [/color]$_5C_1\,=\,5\text{ ways to choose one Other}$
[color=beige]. . [/color]$\text{Hence: }\:1\,\times\,5\:=\:5\text{ ways to have both O#39;s.}$

Therefore, there are:[color=beige] .[/color]$10\,+\,20\,+\,5\:=\:35\text{ possible outcomes.}$

$\text{Your answer of }_7C_3 \,=\,35\text{ is correct!}$

Quote:
 Find the probability that the selection a) Does not contain letter O b) Contains both of the letter O

We've already found the required quantities.

$a)\;P(\text{no O's}) \:=\:\frac{10}{35} \:=\:\frac{2}{7}$

$b)\;P(\text{both O's}) \:=\:\frac{5}{35} \:=\:\frac{1}{7}$

 June 12th, 2011, 03:12 PM #5 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Re: Probability Actually their answer is RIGHT! 35 is the answer only if one of the O's is a capital O and the other O is lower case. Or maybe one of the O's is red and the other is blue. If, on the other hand, the two O's are indistinguishable then the correct answer is 25.
June 12th, 2011, 03:54 PM   #6
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Re: Probability

Quote:
 Originally Posted by mrtwhs Actually their answer is RIGHT! 35 is the answer only if one of the O's is a capital O and the other O is lower case. Or maybe one of the O's is red and the other is blue. If, on the other hand, the two O's are indistinguishable then the correct answer is 25.
This is assuming that the order is NOT important. That is BLG is the same as LBG or GLB or ...

If order is important, then you need to use an exponential generating function and the answer is 135 ways.

 June 12th, 2011, 04:33 PM #7 Senior Member     Joined: Feb 2010 Posts: 711 Thanks: 147 Re: Probability Here are the 25 ways assuming that order is not important. BIL,BIG,BIY,BLG,BLY,BGY,ILG,ILY,IGY,LGY BIO,BLO,BGO,BYO,ILO,IGO,IYO,LGO,LYO,GYO BOO,IOO,LOO,GOO,YOO If order is important then each pattern in the first two rows can be done in 6 ways (6 x 20 = 120) and the five in the last row can each be done in 3 ways (5 x 3 = 15) for a total of 135 ways.
June 12th, 2011, 07:14 PM   #8
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Re: Probability

Quote:
 Originally Posted by mrtwhs Actually their answer is RIGHT! 35 is the answer only if one of the O's is a capital O and the other O is lower case. Or maybe one of the O's is red and the other is blue. If, on the other hand, the two O's are indistinguishable then the correct answer is 25.
I agree with Soroban's response (and I also used the same method to get a & b). The two O's do not have to be visually or distinctually different - they are different. They are two different O's whether they look alike or not. Picking one of them is not the same as picking the other of them. There are two different ways to get an O.

Quote:
 BIO,BLO,BGO,BYO,ILO,IGO,IYO,LGO,LYO,GYO
Each of these has two ways to do them, because each O could be the first one or the second one, and that's 2 different ways. This is where your missing 10 is.

This is a common misconception in probabilities - it's a bit like the fact that a heads on one coin isn't the same as the heads on the other coin, even though they look alike.

(And since the problem did not specify any type of order, I'm assuming order doesn't matter. But even if it did, it's still two different O's that could then have an order to them!)

June 12th, 2011, 08:35 PM   #9
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Re: Probability

Hello, mrtwhs!

Quote:
 Actually their answer is RIGHT! 35 is the answer only if one of the O's is a capital O and the other O is lower case. Or maybe one of the O's is red and the other is blue. If, on the other hand, the two O's are indistinguishable then the correct answer is 25.

The objects do not have to be distinguishable.

Suppose a bowl contains 999 white marbles and 1 black marble.
We draw one marble at random.
What is the probability that it is black?

According to your reasoning, the 999 white balls are identical
[color=beige]. . [/color]so there are two outcomes: (1) the ball is white, (2) the ball is black.

$\text{Therefore: }\:P(\text{black}) \:=\:\frac{1}{2}$

June 13th, 2011, 02:41 AM   #10
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Re: Probability

Quote:
 Originally Posted by soroban Suppose a bowl contains 999 white marbles and 1 black marble. We draw one marble at random. What is the probability that it is black? According to your reasoning, the 999 white balls are identical [color=beige]. . [/color]so there are two outcomes: (1) the ball is white, (2) the ball is black.
and according to your reasoning the probability that it is white is 1/1000. The white ball you picked might not be the one I was looking at.

You have four scrabble tiles N O O N. How many ways can you place them on your tray?

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