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Anna11 [10]
3 years ago
14

How many different 5​-letter radio station call letters can be made a. if the first letter must be Upper C comma Upper X comma U

pper T comma or Upper M and no letter may be​ repeated? b. if repeats are allowed​ (but the first letter is Upper C comma Upper X comma Upper T comma or Upper M​)? c. How many of the 5​-letter radio station call letters​ (starting with Upper C comma Upper X comma Upper T comma or Upper M​) have no repeats and end with the letter Upper S​?
Mathematics
1 answer:
Lady_Fox [76]3 years ago
3 0

Answer:

a) 1,518,000

b) 2,284,880

c) 60,720

Step-by-step explanation:

a) a. if the first letter must be Upper C comma Upper X comma Upper T comma or Upper M and no letter may be​ repeated?

We draw 5 boxes, and based on that we will see the total possible cases. There are 26 alphabets

The first box should have C or X or T or M .No letter may be repeated.

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   25 alphabets   24 alphabets   23 alphabets   22 alphabets

Therefore; total possible call letters = 5 × 25 × 24 × 23 × 22 = 1,518,000

b)

The first box should have C or X or T or M Repeats as allowed

Any                    Any             Any                 Any                    Any

5 alphabets    of the            of the              of the                 of the

C,X , T , M      remaining     remaining       remaining          remaining

                   26 alphabets   26 alphabets   26 alphabets   26 alphabets

Therefore Total possible call letters = 5 × 26 × 26 × 26 × 26 = 2,284,880

c)   The first box should have   C,X , T , M  and end with S

So the last place if fixed, and we now have 25 alphabets. The first box can go in 5 ways. The next box then will have only 24 letters to choose from, as the first box has taken a letter and the last box already has S in it. Repetition not allowed

Any                    Any             Any                 Any                       S

5 alphabets    of the            of the              of the                  is fixed

C,X , T , M      remaining     remaining       remaining           here

                   24 alphabets   23 alphabets   22 alphabets  

Therefore Total possible call letters = 5 × 24 × 23 × 22  × 1 =  60,720

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Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people
Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

p = 0.85, n = 6.

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

P(X < 6) + P(X = 6) = 1

P(X < 6) = 1 - P(X = 6)

In which:

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

So

P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

P(X \leq 4) + P(X > 4) = 1

P(X \leq 4) = 1 - P(X > 4)

In which

P(X > 4) = P(X = 5) + P(X = 6)

P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764

P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

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