Question 1182323
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Part (i)


n = 8 = number of trials
p = 0.4 = probability of success
k = 3 = number of successes that we want


Binomial probability formula
P(k) = (n C k)*(p)^k*(1-p)^(n-k)
P(3) = (8 C 3)*(0.4)^3*(1-0.4)^(8-3)
P(3) = 56*(0.4)^3*(1-0.4)^(8-3)
P(3) = <font color=red>0.27869184</font>
This reprsents the probability of getting exactly 3 successes. The decimal value is exact.


Note: the n C k refers to the combination formula *[Tex \Large _n C _r = \frac{n!}{r!*(n-r)!}] but you replace r with k and you get the same idea.


<font color=red>Answer: 0.27869184</font> (round this however you need to)


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Part (ii)


Use the formula from part (i) to find the following
P(0) = 0.000262144
P(1) = 0.003538944
P(2) = 0.021233664
where n = 9 and p = 0.6 this time (k will range from 0 to 2 as shown above)


Add up those results to get
0.000262144 + 0.003538944 + 0.021233664 = <font color=red>0.025034752</font> 


This represents the probability of getting 2 or fewer (ie at most 2) successes. This decimal value is exact.


<font color=red>Answer: 0.025034752</font> (round this however you need to)
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