Question #346997

Question 1.9 [2, 2, 3]


Chantelle has decided to sell baked biscuits to assist in the payment of her university fees. After


baking for hours and packing packets to sell, she finds that she has 9 biscuits left over. Of these 9


biscuits, 4 are chocolate biscuits, 3 are raisin and 2 are peanut butter. She thinks to herself that she


is going to use these 9 biscuits to assist her with understanding probability. She treats each biscuit


as being slightly different, however order of her selection is not important.


Suppose Chantelle selects 3 biscuits at random from the 9, help her answer the following questions:


a) Calculate the probability that of the 3 biscuits randomly selected, 1 is chocolate, 1 is raisin


and 1 is peanut butter.


b) Calculate the probability that only chocolate biscuits are selected


c) Calculate the probability that at least

1
Expert's answer
2022-06-06T16:59:15-0400

a)


P(3 different)=(41)(31)(21)(93)P(3\ different)=\dfrac{\dbinom{4}{1}\dbinom{3}{1}\dbinom{2}{1}}{\dbinom{9}{3}}

=4(3)(2)84=27=\dfrac{4(3)(2)}{84}=\dfrac{2}{7}

b)


P(3 chocolate)=(43)(30)(20)(93)P(3\ chocolate)=\dfrac{\dbinom{4}{3}\dbinom{3}{0}\dbinom{2}{0}}{\dbinom{9}{3}}

=4(1)(1)84=121=\dfrac{4(1)(1)}{84}=\dfrac{1}{21}

c)


P(at least 1 chocolate)=1P(No chocolate)P(at\ least\ 1\ chocolate)=1-P(No\ chocolate)

=1(40)(3+23)(93)=1-\dfrac{\dbinom{4}{0}\dbinom{3+2}{3}}{\dbinom{9}{3}}

=11(10)84=3742=1-\dfrac{1(10)}{84}=\dfrac{37}{42}


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