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In an experiment a group of children each repeatedly throw a dart at a target - Edexcel - A-Level Maths Statistics - Question 3 - 2018 - Paper 2

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In an experiment a group of children each repeatedly throw a dart at a target. For each child, the random variable H represents the number of times the dart hits th... show full transcript

Worked Solution & Example Answer:In an experiment a group of children each repeatedly throw a dart at a target - Edexcel - A-Level Maths Statistics - Question 3 - 2018 - Paper 2

Step 1

State two assumptions Peta needs to make to use her model.

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Answer

  1. The probability of a dart hitting the target is constant (fixed) for each child and each throw.
  2. The throws of each of the darts are independent, meaning that the outcome of one throw does not affect the others.

Step 2

Using Peta's model, find P(H > 4)

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Answer

To find P(H > 4), we can use the complement rule:

P(H>4)=1P(H4)P(H > 4) = 1 - P(H \leq 4)

Using the Binomial distribution B(10, 0.1):

P(H4)=P(H=0)+P(H=1)+P(H=2)+P(H=3)+P(H=4)P(H \leq 4) = P(H = 0) + P(H = 1) + P(H = 2) + P(H = 3) + P(H = 4)

Calculating separately:

P(H=k)=(10k)(0.1)k(0.9)10kP(H = k) = \binom{10}{k} (0.1)^k (0.9)^{10-k}

Substituting the values:

  • For k = 0: P(H=0)=(100)(0.1)0(0.9)10=0.3487P(H = 0) = \binom{10}{0}(0.1)^0(0.9)^{10} = 0.3487
  • For k = 1: P(H=1)=(101)(0.1)1(0.9)9=0.3874P(H = 1) = \binom{10}{1}(0.1)^1(0.9)^{9} = 0.3874
  • For k = 2: P(H=2)=(102)(0.1)2(0.9)8=0.1937P(H = 2) = \binom{10}{2}(0.1)^2(0.9)^{8} = 0.1937
  • For k = 3: P(H=3)=(103)(0.1)3(0.9)7=0.0574P(H = 3) = \binom{10}{3}(0.1)^3(0.9)^{7} = 0.0574
  • For k = 4: P(H=4)=(104)(0.1)4(0.9)6=0.0119P(H = 4) = \binom{10}{4}(0.1)^4(0.9)^{6} = 0.0119

So,

P(H4)=0.3487+0.3874+0.1937+0.0574+0.0119=0.9981P(H \leq 4) = 0.3487 + 0.3874 + 0.1937 + 0.0574 + 0.0119 = 0.9981

Thus,

P(H>4)=10.9981=0.0019P(H > 4) = 1 - 0.9981 = 0.0019

Step 3

find P(F = 5)

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Answer

Using Peta's model, we need to find the probability of hitting the target for the first time at the fifth throw:

P(F=5)=P(H<5)×P(H=5)P(F=5)= P(H<5) \times P(H=5)

Calculating;

P(F=5)=P(H=0)×P(H=1)×P(H=2)×P(H=3)×P(H=4)×P(H=5)P(F = 5) = P(H = 0) \times P(H = 1) \times P(H = 2) \times P(H = 3) \times P(H = 4) \times P(H = 5)

Using previous calculations, we find:

P(F=5)=(0.3487)(0.3874)(0.1937)(0.0574)(0.0119)=0.0656P(F = 5) = (0.3487)(0.3874)(0.1937)(0.0574)(0.0119) = 0.0656

Step 4

Find the value of α

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Answer

To find α, we first understand that the total probability must sum to 1:

0.01+(11)×α+(21)×α+...+(101)×α=10.01 + (1-1)\times \alpha + (2-1)\times \alpha + ... + (10-1)\times \alpha = 1

Thus, we can write:

0.01+9α=10.01 + 9\alpha = 1

Rearranging gives:

9α=0.999\alpha = 0.99

Thus,

α=0.999=0.11\alpha = \frac{0.99}{9} = 0.11

Step 5

Using Thomas' model, find P(F = 5)

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Answer

Using Thomas' model, we have:

P(F=5)=0.01+(51)×αP(F=5) = 0.01 + (5-1)\times \alpha

Substituting the value of α = 0.11:

P(F=5)=0.01+4×0.11P(F=5) = 0.01 + 4\times 0.11

This simplifies to:

P(F=5)=0.01+0.44=0.45P(F=5) = 0.01 + 0.44 = 0.45

Step 6

Explain how Peta's and Thomas' models differ in describing the probability that a dart hits the target in this experiment.

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Answer

Peta's model assumes a fixed probability of hitting the target for each throw, modeling it as a binomial distribution where the outcome of each throw is independent.

In contrast, Thomas' model introduces a linear progression where the probability adjusts based on the number of previous attempts but caps the maximum throws at 10. This suggests a differing perspective; Peta's approach is based purely on consistent probability, while Thomas's takes into account the number of attempts and introduces a parameter (α) to denote the probability of a successful throw on later attempts.

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