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Terence owns a local shop - AQA - A-Level Maths Pure - 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 Pure - 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 use the property that the total probability must sum to 1:

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

Let's calculate:

  • For n=1n = 1:
    P(N=1)=34(14)11=34P(N = 1) = \frac{3}{4} \left( \frac{1}{4} \right)^{1-1} = \frac{3}{4}

  • For n=2n = 2:
    P(N=2)=34(14)21=316P(N = 2) = \frac{3}{4} \left( \frac{1}{4} \right)^{2-1} = \frac{3}{16}

Now substituting these into the total:

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

To combine the fractions: 1216+316+k=1\frac{12}{16} + \frac{3}{16} + k = 1
1516+k=1\frac{15}{16} + k = 1
Thus,
k=11516=116k = 1 - \frac{15}{16} = \frac{1}{16}

Step 2

Find the probability that a customer, visiting Terence’s shop during the spring, will find at least 2 checkouts staffed.

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Answer

To find the probability that at least 2 checkouts are staffed, we calculate: P(N2)=P(N=2)+P(N=3)P(N \geq 2) = P(N = 2) + P(N = 3)

From part (a), we established:

  • P(N=2)=316P(N = 2) = \frac{3}{16}
  • For n=3n = 3:
    P(N=3)=k=116P(N = 3) = k = \frac{1}{16}

Now, substitute these values: P(N2)=316+116=416=14P(N \geq 2) = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4}

Thus, the final answer is: P(N2)=14P(N \geq 2) = \frac{1}{4}

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