I need help with these probability problems.
1). Five chips are selected from a bag without replacement. The bag originally contained 6 yellow chips and 8 red chips. In how many ways can you choose 5 chips from the bag?
I did 14 C 5 and got 2002, is that correct.
2). In how many ways can you choose no yellow chips?
no
you have to look at the results,
you have 5 chips selected, that could be one of the following cases :
0Y 5R .... RRRRR, only 1 way OR 5!/0!5!)
1Y 4R .... YRRRR, OR RYRRR, ... 5!/4! = 5
2Y 3R .... 5!/(2!3!) = 10
3Y 2R .... 5!/(3!2!) = 10
4Y 1R .... 5!/(4!1!) = 5
5Y OR .... 5!/(5!0!) = 1
total number of ways is 32
This is not a probability question, but rather based on the little formula
for the number of ways that you can arrange p things, q alike of one kind, and r alike of another kind, which is
p!/(q!r!)
2) the number of ways you can choose no yellow chips is 1, namely RRRRR
Had you asked "what is the probability of choosing no yellow chip in choosing any 5 chips from the above that would be
1/( 14C5 ) = 1/2002
Correction:
I said at the end
"Had you asked "what is the probability of choosing no yellow chip in choosing any 5 chips from the above that would be
1/( 14C5 ) = 1/2002 "
that should have been 1/32
1) To find the number of ways to choose 5 chips from the bag without replacement, you need to use the combination formula. The formula for the number of combinations of choosing k objects from a set of n objects without replacement is given by:
n C k = n! / (k!(n-k)!)
Here, n is the total number of chips (14), and k is the number of chips you want to choose (5). Plugging these values into the formula:
14 C 5 = 14! / (5!(14-5)!) = 2002
So yes, you are correct, there are 2002 ways to choose 5 chips from the bag.
2) To find the number of ways to choose no yellow chips, you need to consider that you have only red chips to choose from. Since there are 8 red chips in the bag, you can choose any number of red chips from 0 to 8.
Therefore, there are 8 ways to choose no yellow chips.
To answer the first question, you correctly used the formula for combinations (denoted as nCr or C) to find the number of ways to choose 5 chips from a bag containing 6 yellow chips and 8 red chips without replacement.
The formula for combinations is nCr = n! / (r!(n-r)!), where n represents the total number of chips and r represents the number of chips being chosen.
In this case, you have a total of 14 chips (6 yellow + 8 red) and you want to choose 5 chips, so the formula becomes 14C5 = 14! / (5!(14-5)!).
Simplifying the expression:
14! / (5!(14-5)!) = 14! / (5!9!)
Now, calculating the values:
14! = 14 x 13 x 12 x 11 x 10 x 9!
5! = 5 x 4 x 3 x 2 x 1
Plugging in these values into the formula:
14! / (5!9!) = (14 x 13 x 12 x 11 x 10 x 9!) / (5 x 4 x 3 x 2 x 1 x 9!)
The factor of 9! cancels out in the numerator and denominator:
= (14 x 13 x 12 x 11 x 10) / (5 x 4 x 3 x 2 x 1)
= 2002
Therefore, the number of ways to choose 5 chips from the bag is indeed 2002. Your answer is correct.
Now, let's move on to the second question.
2) In how many ways can you choose no yellow chips?
To determine this, you need to find the combinations of choosing 5 chips from the 8 red chips only. Since you are not choosing any of the 6 yellow chips, the total number of chips to choose from is reduced to 8.
Using the same formula as before, you can calculate it as 8C5 = 8! / (5!(8-5)!)
Simplifying the expression:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
5! = 5 x 4 x 3 x 2 x 1
Plugging in these values into the formula:
8! / (5!3!) = (8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) / (5 x 4 x 3 x 2 x 1 x 3 x 2 x 1)
The factors of 5! and 3! cancel out in the numerator and denominator:
= (8 x 7 x 6) / (3 x 2 x 1)
= 8 x 7 x 6
= 336
Therefore, there are 336 ways to choose no yellow chips from the bag.