Question

1. without using calculator, evaluate $\frac{3}{2 - sqrt{3}}-\frac{2}{3+sqrt{3}}$ leaving your answer in the form $m + nsqrt{3}$
2. an objective test contains 10 questions each with 4 possible options of which one is correct. find to 4 decimal places, the probability that a candidate who guesses the answers get

(a) at most 2 questions correct
(b) less than 1 question correct
(c) greater than 40% of the questions correct.

Explanation:

Question 5

Step1: Rationalize the denominators

Rationalize $\frac{3}{2 - \sqrt{3}}$ by multiplying numerator and denominator by $2+\sqrt{3}$: $\frac{3(2 + \sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{6 + 3\sqrt{3}}{4-3}=6 + 3\sqrt{3}$.
Rationalize $\frac{2}{3+\sqrt{3}}$ by multiplying numerator and denominator by $3-\sqrt{3}$: $\frac{2(3-\sqrt{3})}{(3 + \sqrt{3})(3-\sqrt{3})}=\frac{6-2\sqrt{3}}{9 - 3}=\frac{6-2\sqrt{3}}{6}=1-\frac{1}{3}\sqrt{3}$.

Step2: Subtract the two rational - ized expressions

$(6 + 3\sqrt{3})-(1-\frac{1}{3}\sqrt{3})=(6 - 1)+(3+\frac{1}{3})\sqrt{3}=5+\frac{10}{3}\sqrt{3}$.

Answer:

$5+\frac{10}{3}\sqrt{3}$

Question 6

This is a binomial probability problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 10$ and $p=\frac{1}{4}=0.25$, $1-p = 0.75$.

Part (a)

$P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)$
$P(X=0)=C(10,0)\times(0.25)^{0}\times(0.75)^{10}=\frac{10!}{0!(10 - 0)!}\times1\times(0.75)^{10}=0.0563$
$C(10,1)=\frac{10!}{1!(10 - 1)!}=10$
$P(X = 1)=C(10,1)\times(0.25)^{1}\times(0.75)^{9}=10\times0.25\times(0.75)^{9}=0.1877$
$C(10,2)=\frac{10!}{2!(10 - 2)!}=45$
$P(X = 2)=C(10,2)\times(0.25)^{2}\times(0.75)^{8}=45\times0.0625\times(0.75)^{8}=0.2816$
$P(X\leq2)=0.0563 + 0.1877+0.2816=0.5256$

Part (b)

$P(X\lt1)=P(X = 0)$
$P(X = 0)=C(10,0)\times(0.25)^{0}\times(0.75)^{10}=0.0563$

Part (c)

Greater than $40\%$ of the questions correct means $k>4$, so $P(X>4)=1 - P(X\leq4)$
$P(X = 3)=C(10,3)\times(0.25)^{3}\times(0.75)^{7}=\frac{10!}{3!(10 - 3)!}\times0.015625\times(0.75)^{7}=0.2503$
$P(X = 4)=C(10,4)\times(0.25)^{4}\times(0.75)^{6}=\frac{10!}{4!(10 - 4)!}\times0.00390625\times(0.75)^{6}=0.1460$
$P(X\leq4)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)+P(X = 4)$
$P(X\leq4)=0.0563+0.1877 + 0.2816+0.2503+0.1460=0.9219$
$P(X>4)=1 - 0.9219=0.0781$