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Shkiper50 [21]
2 years ago
7

Suppose that the probability of a baseball player getting a hit in an at-bat is 0.3089. If the player has 25 at-bats during a we

ek, what's the probability that he gets greater than 9 hits?
Mathematics
1 answer:
klio [65]2 years ago
4 0

Answer:

P(X>9) = 0.3593

Step-by-step explanation:

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=25, p=0.3089)  

The probability mass function for the Binomial distribution is given as:  

P(X)=(nCx)(p)^x (1-p)^{n-x}  

For this case we want this probability:

P(X >9)

And we can use the complement rule like this:

P(X>9) = 1-P(X \leq 8)= 1-[P(X=0) + P(X=1) +....+P(X=8)]And we can find the individual probabilities like this:

P(X=0) =(25C0)(0.3089)^0 (1-0.3089)^{25-0} =0.0000974  

P(X=1) =(25C1)(0.3089)^1 (1-0.3089)^{25-1} =0.0011  

P(X=2) =(25C2)(0.3089)^2 (1-0.3089)^{25-2}=0.00584  

P(X=3) =(25C3)(0.3089)^3 (1-0.3089)^{25-3}= 0.02  

P(X=4) =(25C4)(0.3089)^4 (1-0.3089)^{25-4}=0.049  

P(X=5) =(25C5)(0.3089)^5 (1-0.3089)^{25-5}=0.092  

P(X=6)=(25C6) (0.3089)^6 (1-0.3089)^{25-6} = 0.138

P(X=7) =(25C7)(0.3089)^7 (1-0.3089)^{25-7}=0.167

P(X=8) =(25C8)(0.3089)^8 (1-0.3089)^{25-8}=0.168

And in order to do the operations we can use the following excel code:

"=1-BINOM.DIST(8,25,0.3089,TRUE)"  

And we got:

P(X>9) = 0.3593

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3 years ago
20 different comic books will be distributed to five kids. (a) How many ways are there to distribute the comic books if there ar
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Answer:

a) 21684

b) 15504

Step-by-step explanation:

a) Distribute the comic books as follows

only to 1 kid          one way   but we have five kids then  5

to two kids                         C₂₀,₂  =  20! /2! ( 20-2)!    =  20*19 /2  =   190

To three   C₂₀,₃   =  20! /3! ( 20 - 3)!       =  20*19*18 /6               =  1140

To four     C₂₀,₄    =   20 ! / 4! ( 20 - 4 )!   = 20*19*18*17 /4*3*2   =  4845

To five     C₂₀,₅    =    20!  /5! (20 - 5 )!    = 20*19*18*17*16/5*4*3*2*1

C₂₀,₅    =  15504

Then total ways of distribution are:

5 + 190 + 1140 + 4845 + 15504   =  21684

b)   C₂₀,₅  we know from a    that is equal to  15504

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Answer:

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Step-by-step explanation:

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- 16 + 8

- 8

4 0
2 years ago
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Answer:

Step-by-step explanation:

a. 1 & 2

b. 3 & 5

c. 2/15

d. 2/5 x 5/3

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