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Terence owns a local shop - AQA - A-Level Maths Mechanics - Question 11 - 2019 - Paper 3

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Terence owns a local shop. His shop has three checkouts, at least one of which is always staffed. A regular customer observed that the probability distribution for ... show full transcript

Worked Solution & Example Answer:Terence owns a local shop - AQA - A-Level Maths Mechanics - Question 11 - 2019 - Paper 3

Step 1

Find the value of $k$

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Answer

To find the value of kk, we need to ensure that the total probability sums up to 1. We have:

P(N=1)+P(N=2)+P(N=3)=1P(N = 1) + P(N = 2) + P(N = 3) = 1

Substituting the given probabilities:

34(14)0+34(14)1+k=1\frac{3}{4} \left( \frac{1}{4} \right)^0 + \frac{3}{4} \left( \frac{1}{4} \right)^1 + k = 1

Calculating the values:

34+316+k=1\frac{3}{4} + \frac{3}{16} + k = 1

Now turning all terms to fractions having a common denominator of 16:

1216+316+k=1\frac{12}{16} + \frac{3}{16} + k = 1

Thus:

1516+k=1\frac{15}{16} + k = 1

Rearranging gives:

k=11516=116k = 1 - \frac{15}{16} = \frac{1}{16}

Step 2

Select relevant probability for $N \geq 2$ checkouts staffed

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Answer

We need to find the probability that at least 2 checkouts are staffed. This is given by:

P(N2)=P(N=2)+P(N=3)P(N \geq 2) = P(N = 2) + P(N = 3)

Using our earlier results:

P(N=2)=34(14)1=316P(N = 2) = \frac{3}{4} \left( \frac{1}{4} \right)^1 = \frac{3}{16}

And since we found that k=116k = \frac{1}{16} for P(N=3)P(N = 3), we have:

P(N=3)=k=116P(N = 3) = k = \frac{1}{16}

Thus combining both:

P(N2)=316+116=416=14P(N \geq 2) = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}

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