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19 Kirsty either travels by bus or walks when she visits the shops - OCR - GCSE Maths - Question 18 - 2017 - Paper 1

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19 Kirsty either travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.3. The probability that she catches th... show full transcript

Worked Solution & Example Answer:19 Kirsty either travels by bus or walks when she visits the shops - OCR - GCSE Maths - Question 18 - 2017 - Paper 1

Step 1

Complete the tree diagram.

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Answer

To complete the tree diagram, we need to fill in the probabilities for the branches.

  1. For the journey to the shops:

    • The probability of catching the bus is given as 0.3. Therefore, the probability of walking is:

    P(Walking)=1P(Bus)=10.3=0.7P(Walking) = 1 - P(Bus) = 1 - 0.3 = 0.7

  2. For the journey from the shops:

    • The probability of catching the bus is given as 0.8. Therefore, the probability of walking is:

    P(Walking)=1P(Bus)=10.8=0.2P(Walking) = 1 - P(Bus) = 1 - 0.8 = 0.2

The completed tree diagram probabilities are:

  • To the shops:
    • Bus: 0.3
    • Walks: 0.7
  • From the shops:
    • Bus: 0.8
    • Walks: 0.2

Step 2

Show that the probability that Kirsty walks at least one way is 0.76.

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Answer

To find the probability that Kirsty walks at least one way, we can find the probability of her walking both ways and then subtract that from 1 or calculate the individual scenarios in which she walks.

  1. Walking to the shops and walking back:

    • The probability of walking to the shops is 0.7
    • The probability of walking back is 0.2

    The combined probability is:

    P(Walkingextbothways)=0.7imes0.2=0.14P(Walking ext{ both ways}) = 0.7 imes 0.2 = 0.14

  2. Calculate the probability of walking at least one way:
    The probability of NOT walking (i.e. catching the bus both ways) is:

    • Catching the bus to the shops is 0.3
    • Catching the bus back from the shops is 0.8

    Thus, the combined probability of catching the bus both ways is:

    P(Busextbothways)=0.3imes0.8=0.24P(Bus ext{ both ways}) = 0.3 imes 0.8 = 0.24

  3. Final Calculation:
    The probability of walking at least one way is:

    P(Walkingextatleastoneway)=1P(Busextbothways)P(Walking ext{ at least one way}) = 1 - P(Bus ext{ both ways})
    =10.24=0.76= 1 - 0.24 = 0.76

Thus, the probability that Kirsty walks at least one way is confirmed to be 0.76.

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