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There are five matches on each weekend of a football season - HSC - SSCE Mathematics Extension 1 - Question 6 - 2005 - Paper 1
Question 6
There are five matches on each weekend of a football season. Megan takes part in a competition in which she earns one point if she picks more than half of the winnin... show full transcript
Worked Solution & Example Answer:There are five matches on each weekend of a football season - HSC - SSCE Mathematics Extension 1 - Question 6 - 2005 - Paper 1
Step 1
Show that the probability that Megan earns one point for a given weekend is 0.7901, correct to four decimal places.
Answer
To find the probability that Megan earns one point, we first need to determine how many matches she must win. Since she must pick more than half of the winning teams in five matches, she must win at least 3 matches. We can model this situation using the binomial distribution:
Let ( X \sim B(n=5, p=\frac{2}{3}) ). We need to find:
[ P(X \geq 3) = P(X=3) + P(X=4) + P(X=5) ]
Using the binomial probability formula:
[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} ]
we can calculate:
- For ( k=3 ):
[ P(X=3) = \binom{5}{3} \left(\frac{2}{3}\right)^3 \left(\frac{1}{3}\right)^2 = 10 \times \frac{8}{27} \times \frac{1}{9} = \frac{80}{243} ]\n- For ( k=4 ):
[ P(X=4) = \binom{5}{4} \left(\frac{2}{3}\right)^4 \left(\frac{1}{3}\right)^1 = 5 \times \frac{16}{81} \times \frac{1}{3} = \frac{80}{243} ]- For ( k=5 ):
[ P(X=5) = \binom{5}{5} \left(\frac{2}{3}\right)^5 = 1 \times \frac{32}{243} = \frac{32}{243} ]
Thus, combining these:
[ P(X \geq 3) = \frac{80}{243} + \frac{80}{243} + \frac{32}{243} = \frac{192}{243} \approx 0.7901 ]
Step 2
Hence find the probability that Megan earns one point every week of the eighteen-week season. Give your answer correct to two decimal places.
Answer
Since the events are independent, the probability that Megan earns one point each week is the probability of earning a point in one weekend raised to the power of the number of weeks:
[ P(\text{one point every week}) = (0.7901)^{18} \approx 0.0647 ] Thus, rounded to two decimal places, the probability is approximately 0.06.
Step 3
Find the probability that Megan earns at most 16 points during the eighteen-week season. Give your answer correct to two decimal places.
Answer
To find the probability that Megan earns at most 16 points, we can use the complementary probability:
[ P(X \leq 16) = 1 - P(X = 17) - P(X = 18) ]
Calculating for ( P(X=17) ) and ( P(X=18) ) using the binomial formula:
- For ( k=17 ):
[ P(X=17) = \binom{18}{17} (0.7901)^{17} (0.2099)^{1} = 18 \times (0.7901)^{17} \times (0.2099) \approx 0.205 ]\ - For ( k=18 ):
[ P(X=18) = \binom{18}{18} (0.7901)^{18} = (0.7901)^{18} \approx 0.0647 ]
Combined together gives: [ P(X \geq 17) = P(X=17) + P(X=18) \approx 0.205 + 0.0647 = 0.2697 ] Thus: [ P(X \leq 16) = 1 - P(X \geq 17) \approx 1 - 0.2697 = 0.7303 ] Therefore the final answer is approximately 0.73.
Step 4
What is the maximum height the rocket will reach, and when will it reach this height?
Answer
To find the maximum height reached by the rocket, we need to first find the time at which the vertical component of the velocity equals zero. The vertical displacement equation is:
[ y = -4.9t^2 + 200t + 5000 ]
Taking the derivative to find the maximum height:
[ \frac{dy}{dt} = -9.8t + 200 ]
Setting this equal to zero:
[ -9.8t + 200 = 0 \Rightarrow t = \frac{200}{9.8} \approx 20.41 \text{s} ]
Now substituting back to find the maximum height:
[ y_{max} = -4.9(20.41)^2 + 200(20.41) + 5000 \approx 5450.10 \text{ m} ]
Step 5
The pilot can only operate the ejection seat when the rocket is descending at an angle between 45° and 60° to the horizontal. What are the earliest and latest times that the pilot can operate the ejection seat?
Answer
The angle of descent can be determined by the ratio of vertical and horizontal components of velocity. The equations for velocity are: [ \frac{dy}{dt} = -9.8t + 200 ] and [ \frac{dx}{dt} = 200 ] The angle ( \theta ) is given by ( \tan(\theta) = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} ). We require: [ an(45°) \leq \frac{-9.8t + 200}{200} \leq an(60°) ] Calculating these inequalities gives the required times for ejection.
Step 6
For the parachute to open safely, the pilot must eject when the speed of the rocket is no more than 350 m/s. What is the latest time at which the pilot can eject safely?
Answer
The speed of the rocket is given by:
[ v = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} = \sqrt{(200)^2 + (-9.8t + 200)^2} ]
Setting this to be less than or equal to 350 m/s:
[ \sqrt{40000 + (-9.8t + 200)^2} \leq 350 ]
Solving this inequality gives the latest time for safe ejection.