Question Synopsis: A survey found that 23% of respondents stick with their tour group. (No total number of travelers was given).
In a sample of 6 travelers, what is the probability that two will stick with their group?
In a sample of 6 travelers, what is the probability that AT LEAST two will stick with their group?
In a sample of 10 travelers, what is the probability that NONE will stick with their group?
Any help on determining the answers will be very appreciated! Thanks in advance!
Answer In a sample of 10 travelers, what is the probability that NONE will stick with their group?
That means
Trav 1 does not stick to his tour group AND Trav 2 does not stick to his tour group ...
So, it is 0.77 * 0.77 * ...
i.e. 0.77 to the power of 10.
In a sample of 6 travelers, what is the probability that two will stick with their group?
Suppose
Trav 1 sticks to his tour group AND Trav 2 sticks to his tour group
and Trav 3 does not stick to his tour group AND Trav 4 does not stick to his tour group ...
i,e, the chance is
0.23 * 0.23 * 077 * 0.77 * ...4 times
i.e. 0.23 to the power 2 * 0.77 to the power of 4
But this may happen in many ways:
it may be Trav 1,2 or Trav 1,3 or Trav 1,4 or ...
How many different ways are there?
There 6 ways of choosing 1 number out of 6, 5 ways of choosing a second number; but there is overlap ( it is same if I choose 1,5 and 5,1)
So, total number of ways is
6*5/2
i.e. the chance of exactly 2 stciking-to-tour guide is
i.e. 0.23 to the power 2 * 0.77 to the power of 4
Here is a similar problem/solution:
When Susan and Jessica play a card game, Susan wins 60% of the time. If they play 9 games, what is the probability that Jessica will have won more games than Susan?
Answer: If they play 9 games, what is the probability that Jessica will have won more games than Susan?
That means that
JEssica wins 5 or more games
i.e JEssica wins 5 games OR JEssica wins 6 games OR JEssica wins 7 games OR JEssica wins 8 games OR JEssica wins 9 games
For each of these, let us find the probability.
Then we can add these probabilities.
A) What is the chance that JEssica wins 9 games?
One game: 0.4
All the games, means
she wins the first AND then she wins the second AND then she wins the third...
So,
0.4 * 0.4 * 0.4 * ...
i.e. 0.4 to the power 9
B) What is the chance that JEssica wins 8 out of 9 games?
Suppose JEssica wins the first 8 and then susan wins
that means
J wins the first AND then she wins the second AND then she wins the third...8 times.. AND SUSAN WINS 1 game
So,
0.4 * 0.4 * 0.4 * ...8 times and 0.6
i.e. 0.4 to the power 8 * 0.6
But this can happen in many ways: Susan may win the last game OR the first game or the second...
There are 9 ways this might happen...
So, the chance that JEssica wins 8 out of 9 games..
is 9 * 0.4 to the power 8 * 0.6
C) What is the chance that JEssica wins 7 out of 9 games?
Suppose JEssica wins the first 7 and then susan wins 2 games
that means
J wins the first AND then she wins the second AND then she wins the third...7 times.. AND SUSAN WINS 2 game
So,
0.4 * 0.4 * 0.4 * ...7 times and 0.6 * 0.6
i.e. 0.4 to the power 7 * 0.6 to the power 2
But this can happen in many ways: Susan may win the last 2 games OR the first 2 games or the second and sixth games or...
How many ways can this happen:
i.e. how many ways can you choose 2 numbers out of 9 ?
There are 9 ways you can choose 1 number, then 8 ways you can choose the other.
So, is it 9*8 ways?
( Wait! You might choose 5 first, then 7; or you might choose 7 first, and then 5 .. aren't these the same!)
So, it is 9*8/2 ways?
So, the chance that JEssica wins 7 out of 9 games..
is 9*8/2 * 0.4 to the power 7 * 0.6 to the power 2
D) What is the chance that JEssica wins 6 out of 9 games?
Suppose JEssica wins the first 6 and then susan wins 3 games
that means...
fill in the steps here ...
i.e. 0.4 to the power 6 * 0.6 to the power 3
But this can happen in many ways: Susan may win the last 3 games OR ...
How many ways can this happen:
i.e. how many ways can you choose 3 numbers out of 9 ?
There are 9 ways you can choose 1 number, then 8 ways you can choose the other, 7 ways of choosing the 3rd number
So, is it 9*8*7 ways?
( Wait! You might choose 5 first, then 7, then 1; or you might choose 7 first, and then 5, then 1 .. aren't these the same!)
So, it is 9*8*7/3*2 ways?
So, the chance that JEssica wins 6 out of 9 games..
is 9*8*7/3*2 * 0.4 to the power 6 * 0.6 to the power 3
E) What is the chance that JEssica wins 5 out of 9 games?
fill in the steps here ...
i.e. 0.4 to the power 5 * 0.6 to the power 4
But this can happen in many ways:
fill in the steps
So, it is 9*8*7*6/4*3*2 ways?
So, the chance that JEssica wins 4 out of 9 games..
is 9*8*7*6/4*3*2 * 0.4 to the power 5 * 0.6 to the power 4
We said at first ..
JEssica wins 5 or more games
i.e JEssica wins 5 games OR JEssica wins 6 games OR JEssica wins 7 games OR JEssica wins 8 games OR JEssica wins 9 games
For each of these, let us find the probability.
Then we can add these probabilities.
So, total chance =
i.e. 0.4 to the power 9
+ 9 * 0.4 to the power 8 * 0.6
+ 9*8/2 * 0.4 to the power 7 * 0.6 to the power 2
+ 9*8*7/3*2 * 0.4 to the power 6 * 0.6 to the power 3
+ 9*8*7*6/4*3*2 * 0.4 to the power 5 * 0.6 to the power 4
Notice a pattern ?
Confused?
Want to discuss?
Write to me at surajit@gmail.com
