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Vinvika [58]
3 years ago
12

A pet store contains 35 light green parakeets (14 females and 21 males) and 44 sky blue parakeets (28 females and 16 males). You

randomly choose one of the parakeets. What is the probability that it is a female or a sky blue parakeet?
Mathematics
1 answer:
snow_tiger [21]3 years ago
7 0
You can use this formula <span>P(AorB) = P(A) + P(B) - P(AandB) 

Given:
35 LG (14 F & 21 M)
44 SB (28 F & 16 M)

Req:
- the probability that it is a female (F) or a sky blue (SB)

Sol:
</span>P(F or SB) = P(F) + P(SB) - P(F and SB) 
                 = [(14 F + 28 F)/(35 + 44)] + [(44 SB)/(35 + 44)] - [(28 F)/(35 + 44)]
                 = 53.16 + 55.70 - 35.44
                 = 73.42%

You have to deduct 28 female parakeets from 44 sky blue parakeets because the 28 parakeets are already accounted for in the female parakeets. You can also think of how many ways you can choose a female parakeet and a sky blue parakeet. Add all female parakeets (14 + 28) = 42. Sky blue parakeet equaled to 44. Minus the 28 female parakeets included in the sky blue parakeet to avoid double counting. 42 + 44 - 28 = 58 divided by 79 (35 + 44) total parakeets = 73.42%


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a) X \sim Binom(n=16, p=0.2)  

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

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

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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".  

Part a

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

X \sim Binom(n=16, p=0.2)  

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

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

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For this case we want this probability:

P(X=0)

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

P(X)=(nCx)(p)^x (1-p)^{n-x}  

Where (nCx) means combinatory and it's given by this formula:  

nCx=\frac{n!}{(n-x)! x!}  

And using this function we got:

P(X=0)=(16C0)(0.2)^{0} (1-0.2)^{16-0}=0.02815

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