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Pavlova-9 [17]
3 years ago
9

Suppose Mike has a dresser with 3 tan pants, 5 jeans, 2 black pants, and 3 gray pants. He also has 2 brown shoes, 3 black shoes,

and 2 blue shoes. If he randomly chooses a pair of pants and shoes, what is the probability that both are black? Give your answer as an exact fraction and reduce the fraction as much as possible.
Mathematics
1 answer:
Inessa [10]3 years ago
6 0

Answer:

Probability of choosing a black pair of pants and shoes = \frac{6}{91}

Step-by-step explanation:

Given:

Total number of pants Mike has =\textrm{ 3 tan pants + 5 jeans + 2 black pants + 3 gray pants = 13 pair of pants}

Probability of randomly choosing a pair of black pants = \frac{\textrm{Number of Black pants}}{\textrm{Total number of pants}} = \frac{2}{13}

Total number of shoes Mike has = \textrm{ 2 brown shoes + 3 black shoes + 2 blue shoes = 7 pairs of shoes}

Probability of randomly choosing a pair of black shoes = \frac{\textrm{Number of Black shoes}}{\textrm{Total number of shoes}} = \frac{3}{7}

Probability of choosing a black pair of pants and shoes =  \frac{2}{13}\times \frac{3}{7} = \frac{6}{91}

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The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

We can model this as a binomial random variable, with p=0.57 and n=14.

P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\

P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\

P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\

P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886

7 0
3 years ago
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