Question 1: Conditional Probability (Year 2)
Tisam is playing a game.
She uses a ball, a cup and a spinner.
The random variable $X$ represents the number the spinner lands on when it is spun.
The probability distribution of $X$ is given in the following table
where $a$, $b$, $c$ and $d$ are probabilities.
To play the game
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the spinner is spun to obtain a value of $x$
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Tisam then stands $x\text{ cm}$ from the cup and tries to throw the ball into the cup
The event $S$ represents the event that Tisam successfully throws the ball into the cup.
To model this game Tisam assumes that
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$\text{P}(S \mid \{X = x\}) = \frac{k}{x}$ where $k$ is a constant
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$\text{P}(S \cap \{X = x\})$ should be the same whatever value of $x$ is obtained from the spinner
Using Tisam’s model,
Nav tries, a large number of times, to throw the ball into the cup from a distance of 100 cm.
He successfully gets the ball in the cup 30% of the time.
State, giving a reason, why Tisam’s model of this game is not suitable to describe Nav playing the game for all values of $X$
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Correct Result
(a)
(b) * $a = \boxed{\frac{2}{25}}$ (or $0.08$)
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$b = \boxed{\frac{1}{5}}$ (or $0.20$)
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$c = \boxed{\frac{8}{25}}$ (or $0.32$)
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$d = \boxed{\frac{2}{5}}$ (or $0.40$)
(c) If $\text{P}(S \mid \{X = 100\}) = 0.3$, then $\frac{k}{100} = 0.3 \implies k = 30$.
This model would mean $\text{P}(S \mid \{X = 20\}) = \frac{30}{20} = 1.5$.
Since a probability cannot be greater than $1$, the model is not suitable.
Did you get this correct?
Step-by-Step
Solution for Part A
We are given two key assumptions from Tisam's model:
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$\text{P}(S \mid \{X = x\}) = \frac{k}{x}$
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$\text{P}(S \cap \{X = x\})$ is a constant value for all $x$. Let's call this constant $V$.
We know the general probability rule: $\text{P}(A \cap B) = \text{P}(A) \times \text{P}(B \mid A)$.
Applying this to our variables:
Let's evaluate this for $x = 50$ and $x = 80$:
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For $x = 50$: $\text{P}(S \cap \{X = 50\}) = b \times \frac{k}{50}$
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For $x = 80$: $\text{P}(S \cap \{X = 80\}) = c \times \frac{k}{80}$
Since the intersection is constant, we can equate them:
Divide both sides by $k$ (since it's a constant):
Rearrange to solve for $c$:
Solution for Part B
First, we need to express all probabilities ($a$, $b$, $c$, and $d$) in terms of a single variable, like $b$. We already have $c = \frac{8}{5}b$. Now, do the same for $a$ and $d$.
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For $x = 20$: $a \times \frac{k}{20} = b \times \frac{k}{50} \implies \frac{a}{20} = \frac{b}{50} \implies a = \frac{20}{50}b = \frac{2}{5}b$
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For $x = 100$: $d \times \frac{k}{100} = b \times \frac{k}{50} \implies \frac{d}{100} = \frac{b}{50} \implies d = \frac{100}{50}b = 2b$
The sum of all probabilities in a distribution must equal $1$:
Substitute the expressions we found:
Combine the terms:
Now calculate the actual values for $a$, $c$, and $d$:
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$a = \frac{2}{5}(\frac{1}{5}) = \frac{2}{25}$
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$c = \frac{8}{5}(\frac{1}{5}) = \frac{8}{25}$
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$d = 2(\frac{1}{5}) = \frac{2}{5}$ (which is $\frac{10}{25}$)
Solution for Part C
Nav's experimental data shows that for $x = 100$, the probability of success is $30\%$, or $0.3$.
Using Tisam's model formula $\text{P}(S \mid \{X = x\}) = \frac{k}{x}$, we can find the implied value of $k$ for Nav:
If we apply this constant $k = 30$ to a different distance, say $x = 20$, the model breaks:
Since a probability cannot be greater than $1$, this model is not suitable to describe Nav playing the game for all values of $X$.