Need help with some HSC 3U Maths Questions (1 Viewer)

Leo18032003 · Jun 20, 2021

Hello everyone! Can anybody help me to solve these question? (sorry I'm very bad at Maths, took me forever but I still couldn't figure out

)

1. (HSC 1982): A biased coin is found to have probability 0.7 of showing a head when tossed.
b) What is the most likely combination of heads and tails to occur in a sequence of five tosses? What is the probability that this event occurs? (Ans: 4H,5T)

2. A student is taking a test with 50 multiple choice questions and guesses the answer to each one. The probability of guessing correctly is 0.3.
b) What is the most likely number of question answered correctly? (Ans:15)

3. A die is biased so in any single throw the probability of an odd score is p where p is a constant such that 0<p<1, p≠0.5. Find the probability that in six throws of the die the product of the scores is even. (Ans: 1 - p^6)

4. Show that AF:FB = 4:1

Screen Shot 2021-06-20 at 2.03.29 pm.png

5.

Screen Shot 2021-06-20 at 2.03.40 pm.png

6. Given:

Screen Shot 2021-06-20 at 2.03.51 pm.png

Screen Shot 2021-06-20 at 2.03.56 pm.png

Really appreciate the help!!!!

Everwinter · Jun 20, 2021

This is some very messy stuff but it is the solution to question 4.

Everwinter · Jun 20, 2021

I may try to do other questions later, but these questions need to take a super long time so I can't do it altogether. (definitely not just reluctant to do probabilities lol). If you have a question on question 4, feel free to ask.

Leo18032003 · Jun 20, 2021

Everwinter said:
I may try to do other questions later, but these questions need to take a super long time so I can't do it altogether. (definitely not just reluctant to do probabilities lol). If you have a question on question 4, feel free to ask.

Thank you so much

! I finally got it after seeing your working out! (lol probabilities are nightmares

)

Everwinter · Jun 21, 2021

question 6

Directrix · Jun 21, 2021

I can help with 2.

mean = np (n = number of trials, p = probability of success)
n = 50
p = 0.3

50 * 0.3 = 15 answers correct

that was actually on my test last tuesday

CM_Tutor · Jun 28, 2021

Leo18032003 said:
1. (HSC 1982): A biased coin is found to have probability 0.7 of showing a head when tossed.
b) What is the most likely combination of heads and tails to occur in a sequence of five tosses? What is the probability that this event occurs? (Ans: 4H,5T)

This is a binomial probability problem.

If the probability of a head is $p$ and the probability of a tail is $q=1-p$ , then the probability of $k$ heads in $n$ tosses is:

$P(k\ \text{heads in}\ n\ \text{tosses}) = \binom{n}{k} p^k q^{n-k}$

We know here that $p = 0.7$ and so $q = 0.3$ and we have $n = 5$ tosses. So:

$P(0\ \text{heads in}\ 5\ \text{tosses}) = \binom{5}{0} 0.7^0 \times 0.3^5 = 1 \times 1 \times 0.3^5 = 0.00243$

$P(1\ \text{heads in}\ 5\ \text{tosses}) = \binom{5}{1} 0.7^1 \times 0.3^4 = 5 \times 0.7 \times 0.3^4 = 0.02835$

$P(2\ \text{heads in}\ 5\ \text{tosses}) = \binom{5}{2} 0.7^2 \times 0.3^3 = 10 \times 0.7^2 \times 0.3^3 = 0.1323$

$P(3\ \text{heads in}\ 5\ \text{tosses}) = \binom{5}{3} 0.7^3 \times 0.3^2 = 10 \times 0.7^3 \times 0.3^2 = 0.3087$

$P(4\ \text{heads in}\ 5\ \text{tosses}) = \binom{5}{4} 0.7^4 \times 0.3^1 = 5 \times 0.7^4 \times 0.3 = 0.36015$

$P(5\ \text{heads in}\ 5\ \text{tosses}) = \binom{5}{5} 0.7^5 \times 0.3^0 = 1 \times 0.7^5 \times 1 = 0.16807$

So, the most likely outcome is 4 heads in 5 tosses

Now, if you look at the pattern here, you can see that the probabilities increase with $k$ until a maximum is reached, and then decrease. Note that it is possible that there may be two, equally likely outcomes, rather than a single maximum value. In any case, rather than waste time computing each probability (especially if the coin was tossed 50 times!), it would be preferable to be able to seek $k$ directly. There are two ways this can be done.

Method 1: Use expectations

The most likely outcome for a binomial will be at or near the mean ("near" as the mean may not be an integer).

In the above case, the mean is $E(X) = np = 5 \times 0.7 = 3.5$ and so the outcome with the greatest probability will be either $k = 3$ tosses or $k = 4$ tosses. We then need only compute the probabilities for those two values of $k$ and select the larger. However, you would need to justify this in an exam by showing the computation of the expectation and commenting that the most likely outcome will be at / near this result.

Suppose the question had been 50 tosses, in which case $E(X) = 35$ . The probabilities will follow $P(k\ \text{heads in}\ 50\ \text{tosses}) = \binom{50}{k} 0.7^k \times 0.3^{50-k}$ , and we can examine the results around $k = 35$ :

$P(34\ \text{heads in}\ 50\ \text{tosses}) = \binom{50}{34} 0.7^{34} \times 0.3^{16} = 0.1147...$

$P(35\ \text{heads in}\ 50\ \text{tosses}) = \binom{50}{35} 0.7^{35} \times 0.3^{15} = 0.122346...$

$P(34\ \text{heads in}\ 50\ \text{tosses}) = \binom{50}{36} 0.7^{36} \times 0.3^{14} = 0.1185948...$

We can see that the most likely outcome is 35 heads and 15 tails in 50 tosses of this biased $P(\text{Head}) = p = 0.7$ coin.

Method 2: Use binomials

Again considering the case with 50 tosses where $P(k\ \text{heads in}\ 50\ \text{tosses}) = \binom{50}{k} 0.7^k \times 0.3^{50-k}$ , consider the ratio:

$\begin{align*} R &= \cfrac{P(k+1\ \text{heads in}\ 50\ \text{tosses})}{P(k\ \text{heads in}\ 50\ \text{tosses})} \\ &= \cfrac{\binom{50}{k+1} 0.7^{k+1} \times 0.3^{50-(k+1)}}{\binom{50}{k} 0.7^k \times 0.3^{50-k}} \\ &= \left(\cfrac{50!}{(k+1)!(50-(k+1))!} \div \cfrac{50!}{k!(50-k)!}\right) \times \cfrac{0.7^{k+1}}{0.7^k} \times \cfrac{0.3^{49-k}}{0.3^{50-k}} \\ &= \left(\cfrac{50!}{(k+1)!(49-k)!} \times \cfrac{k!(50-k)!}{50!}\right) \times \cfrac{0.7^{k + 1 - k}}{0.3^{50-k - (49 - k)}} \\ &= \left(\cfrac{50!}{50!} \times \cfrac{k!}{(k+1)!} \times \cfrac{(50-k)!}{(49-k)!}\right) \times \cfrac{0.7}{0.3^{50 - k - 49 + k}} \\ &= \left(1 \times \cfrac{k!}{(k+1) \times k!} \times \cfrac{(50-k) \times (49-k)!}{(49-k)!}\right) \times \cfrac{0.7}{0.3} \\ &= \cfrac{50-k}{k+1} \times \cfrac{7}{3} \\ &= \cfrac{7(50 - k)}{3(k + 1)} \end{align*}$

Now, we can solve the inequality

$\begin{align*} R &> 1 \\ \cfrac{7(50 - k)}{3(k + 1)} &> 1 \\ 7(50 - k) &> 3(k + 1) \quad \text{since $k$ is an integer and $0 \leqslant k \leqslant 49$, thus $3(k + 1) > 0$} \\ 350 - 7k &> 3k + 3 \\ 343 &> 10k \\ k &< 34.3 \end{align*}$

We thus know that, for every $k \leqslant 34$ the value of

$R = \cfrac{P(k+1\ \text{heads in}\ 50\ \text{tosses})}{P(k\ \text{heads in}\ 50\ \text{tosses})} > 1$

and hence

$P(k+1\ \text{heads in}\ 50\ \text{tosses}) > P(k\ \text{heads in}\ 50\ \text{tosses})$

which shows that

$P(0\ \text{heads in}\ 50\ \text{tosses}) < P(1\ \text{heads in}\ 50\ \text{tosses}) < P(2\ \text{heads in}\ 50\ \text{tosses}) \\ < P(3\ \text{heads in}\ 50\ \text{tosses}) < ... < P(34\ \text{heads in}\ 50\ \text{tosses}) < P(35\ \text{heads in}\ 50\ \text{tosses})$

with the last pair coming from using $k = 34$ .

We can easily prove that $R < 1$ for $k \geqslant 35$ , which gives another series of inequalities:

$P(35\ \text{heads in}\ 50\ \text{tosses}) > P(36\ \text{heads in}\ 50\ \text{tosses}) > P(37\ \text{heads in}\ 50\ \text{tosses}) \\ > P(38\ \text{heads in}\ 50\ \text{tosses}) > ... > P(49\ \text{heads in}\ 50\ \text{tosses}) > P(50\ \text{heads in}\ 50\ \text{tosses})$

with the last pair coming from using $k = 49$ .

We have then proved (rigorously) that the pattern of probabilities does indeed increase to a maximum when there are 35 heads seen in the 50 tosses, and then decrease again.

Our conclusion is that the most likely outcome is $P(35\ \text{heads in}\ 50\ \text{tosses}) = \binom{50}{35} 0.7^{35} 0.3^{15} = 0.122346...$ .

CM_Tutor · Jun 28, 2021

Directrix said:
I can help with 2.

mean = np (n = number of trials, p = probability of success)
n = 50
p = 0.3

50 * 0.3 = 15 answers correct

that was actually on my test last tuesday

The answer to question 2 could also be expanded by one of the methods in my post above to actually calculate P(14 / 50) correct, P(15 / 50) correct, and P(16 / 50) correct.

You should find that

P(0 / 50) < P(1 / 50) < ... < P(14 / 50) < P(15 / 50) > P(16 / 50) > P(17 / 50) > ... > P(50 / 50)

Directrix · Jun 28, 2021

CM_Tutor said:
The answer to question 2 could also be expanded by one of the methods in my post above to actually calculate P(14 / 50) correct, P(15 / 50) correct, and P(16 / 50) correct.

You should find that

P(0 / 50) < P(1 / 50) < ... < P(14 / 50) < P(15 / 50) > P(16 / 50) > P(17 / 50) > ... > P(50 / 50)

yeah, that is more logical, but i just wanted to do the more "elementary style"

like ie i expect to get 5 heads when i roll a fair coin ten times rather than the harder calculation

so i just multiplied trials by success probability

CM_Tutor · Jun 28, 2021

Leo18032003 said:
3. A die is biased so in any single throw the probability of an odd score is p where p is a constant such that 0<p<1, p≠0.5. Find the probability that in six throws of the die the product of the scores is even. (Ans: 1 - p^6)

We have a biased six-sided die where P(odd score, i.e. 1, 3, or 5) = $p$ .

The two outcomes, odd and even, are mutually exclusive (that is, it is impossible for both to occur) and complementary (that is, if one does not occur, then it is certain that the other must have occurred). In short, this is a Bernoulli experiment.

Thus, so P(even score, i.e. 2, 4, or 6) = $q = 1 - p$ .

With multiple rolls of the die, we have a binomial situation and so from $n$ rolls of the dice, the probability of $k$ odd scores is

$P(k\ \text{odd scores from}\ n\ \text{rolls}) = \binom{n}{k} p^k q^{n - k} = \binom{n}{k} p^k (1 - p)^{n - k}$

Now, for the product of the scores to be even, at least one of them must be even - otherwise, it will be a product of all odd scores, which is odd - but once there is a single even score, the product must have a factor of 2 and thus must be even. So, the question actually seeks the probability that there is at least one even score.

$\begin{align*} P(\text{At least 1 even score from $6$ rolls}) &= P(\text{$1$ or $2$ or ... or $6$ even scores from $6$ rolls}) \\ &= 1 - P(\text{exactly $0$ even scores from $6$ rolls}) \\ &= 1 - P(\text{exactly $6$ odd scores from $6$ rolls}) \\ &= 1 - \binom{6}{6} p^6 (1 - p)^{6 - 6} \\ &= 1 - 1 \times p^6 \times 1 \qquad \text{as $(1 - p)^0 = 1$} \\ &= 1 - p^6 \end{align*}$

CM_Tutor · Jun 28, 2021

Directrix said:
yeah, that is more logical, but i just wanted to do the more "elementary style"

like ie i expect to get 5 heads when i roll a fair coin ten times rather than the harder calculation

so i just multiplied trials by success probability

Your answer was / is fine as the question asks for a simple fact ("What is the number ...").

I was pointing out for the OP and other readers that a more rigorous approach is available, especially if the question asked for evidence / justification or had a mark weighting that was inconsistent with a simple statement after calculating $E(X)$ .

Directrix · Jun 28, 2021

CM_Tutor said:
Your answer was / is fine as the question asks for a simple fact ("What is the number ...").

I was pointing out for the OP and other readers that a more rigorous approach is available, especially if the question asked for evidence / justification or had a mark weighting that was inconsistent with a simple statement after calculating $E(X)$ .

yeah, better your way, and thanks for explaining, helps all

CM_Tutor · Jun 28, 2021

Directrix said:
yeah, better your way, and thanks for explaining, helps all

If I was answering in an exam question, I would have used your way to get $E(X) = 15$ and then, if support was needed, calculated the three probabilities for P(14 / 50), P(15 / 50), and P(16 / 50). The full binomial proof, which relates to something that was asked regularly under the pre-2020 syllabus, is much less likely to be asked now - not without structure / support, at least.

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