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Probability Questions

chee

New member
Joined
May 3, 2012
Messages
12
Need someone to check if my answers to these probability questions are right :$

Q1: Two jokers are added to a standard 52 card deck. The deck is then shuffled, what is the probability that someone draws 5 cards and two of those 5 cards happen to be both jokers.

is this what you do? [SUB]5[/SUB]C[SUB]3 / 54[/SUB]C[SUB]5 [/SUB]?

Q2: 7 members of the swimming club line up for a picture. What is the probability that Marry and Jory will sit together?

is the answer, 12/7! ?

Q3: There is a 15% of winning a game, what are the odds against winning the game?

is the answer, 17:3?

Q4: From a group of 7 juniors and 10 seniors, determine how many communities of 6 students can be chosen if 4 are juniors?

is the answer, ([SUB]7[/SUB]C[SUB]4[/SUB])([SUB]10[/SUB]C[SUB]2[/SUB]) ?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, chee!

(1) Two Jokers are added to a standard 52 card deck; the deck is then shuffled.,
What is the probability that 5 cards are drawn and two of those 5 cards happen to be Lokers?

Is this what you do? .\(\displaystyle \dfrac{_5C_3}{_{54}C_5}\) . no

I don't understand your numerator . . . choose 3 objects from 5?


There are: .\(\displaystyle _{54}C_5 \,=\,3,\!162,\!510\) possible outcomes.

We want both Jokers . . . There is: .\(\displaystyle _2C_2 \,=\,1\) way.
We want 3 card from the remaining 52: .\(\displaystyle _{52}C_3\,=\,22,\!100\) ways.
. . There are: .\(\displaystyle 1\cdot 22,100 \:=\:22,\!100\) hands with two Jokers.

\(\displaystyle P(\text{2 Jokers}) \:=\:\dfrac{22,\!100}{3,\!162,\!510} \:=\:\dfrac{170}{24,\!327}\)




(2) Seven members of the swimming club line up for a picture.
What is the probability that Mary and Jory will sit together?

Is the answer: .\(\displaystyle \dfrac{12}{7!}\,?\) . no

Again, I don't understand your numerator . . .

There are seven people: .\(\displaystyle A\;B\;C\;D\;E\;M\;J\)
The denominator is \(\displaystyle 7!\)

Duct-tape Mary and Jory together.
Then we have six "people" to arrange: .\(\displaystyle A\;B\;C\;D\;E\;\boxed{MJ}\)
. . They can be arranged in \(\displaystyle 6!\) ways.
But Mary and Jory could be taped as \(\displaystyle \boxed{MJ}\) or \(\displaystyle \boxed{JM}\)
. . So there are \(\displaystyle 2\) choices.
Hence, there are: .\(\displaystyle 2\cdot6!\) ways that Mary and Jory are adjacent.

\(\displaystyle P(\text{Mary and Jory adjacent}) \;=\;\dfrac{2\cdot6!}{7!} \;=\;\dfrac{2}{7}\)



(3) There is a 15% of winning a game, what are the odds against winning the game?

Is the answer \(\displaystyle 17:3\,?\)
Yes!



(4) From a group of 7 juniors and 10 seniors, 6 students are chosen at random.
Determine how many committees can be chosen with 4 juniors and 2 seniors.

Is the answer: .\(\displaystyle (_7C_4)(_{10}C_2)\,?\)
Yes!
 

chee

New member
Joined
May 3, 2012
Messages
12

There are seven people: .\(\displaystyle A\;B\;C\;D\;E\;M\;J\)
The denominator is \(\displaystyle 7!\)
ra
Duct-tape Mary and Jory together.
Then we have six "people" to arrange: .\(\displaystyle A\;B\;C\;D\;E\;\boxed{MJ}\)
. . They can be arranged in \(\displaystyle 6!\) ways.
But Mary and Jory could be taped as \(\displaystyle \boxed{MJ}\) or \(\displaystyle \boxed{JM}\)
. . So there are \(\displaystyle 2\) choices.
Hence, there are: .\(\displaystyle 2\cdot6!\) ways that Mary and Jory are adjacent.

thanks for the help :D
but what if I changed that question to what is the probability that Marry and Jory would randomly sit together?
would anything change? do we still have to "duck tape" them together as one unit or no?

my original reasoning for 12/7! was this,

There are 7! ways for the 7 members to line up. There are six ways for Mac to be directly to the left of Jon and six ways for Mac to be directly to the right of Jon, so there are 12 ways for Mac and Jon to sit together


M J X X X X X
X M J X X X X
X X M J X X X
X X X M J X X
X X X X M J X
X X X X X M J


reverse M and J for the remaining 6 ways.


So the probability is, 12/7! ?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello again, chee!

Thank you for the explanation.

Your reasoning is excellent,
. . but you forgot to arrange the other 5 people.

The numerator would be: .\(\displaystyle 12\!\cdot\!5!\)

And the answer is: .\(\displaystyle \dfrac{12\!\cdot\!5!}{7!} \:=\:\dfrac{2}{7}\)
 
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