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 fractions

$(6 + 3\sqrt{3})-(1-\frac{1}{3}\sqrt{3})=(6 - 1)+(3+\frac{1}{3})\sqrt{3}=5+\frac{9 + 1}{3}\sqrt{3}=5+\frac{10}{3}\sqrt{3}=4+\sqrt{3}$ (after simplification).

Question 6

This is a binomial probability problem where $n = 10$ (number of questions), $p=\frac{1}{4}=0.25$ (probability of getting a question correct by guessing), and $q = 1 - p=0.75$ (probability of getting a question wrong by guessing). The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.

Part (a)

Step1: Calculate $P(X = 0)$, $P(X = 1)$ and $P(X = 2)$

$P(X = 0)=C(10,0)\times(0.25)^{0}\times(0.75)^{10}=\frac{10!}{0!(10 - 0)!}\times1\times(0.75)^{10}=0.0563$.
$P(X = 1)=C(10,1)\times(0.25)^{1}\times(0.75)^{9}=\frac{10!}{1!(10 - 1)!}\times0.25\times(0.75)^{9}=10\times0.25\times0.0751 = 0.1877$.
$P(X = 2)=C(10,2)\times(0.25)^{2}\times(0.75)^{8}=\frac{10!}{2!(10 - 2)!}\times0.0625\times(0.75)^{8}=45\times0.0625\times0.1001=0.2816$.

Step2: Calculate $P(X\leq2)$

$P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)=0.0563+0.1877 + 0.2816=0.5256$.

Part (b)

Step1: Calculate $P(X<1)$

$P(X<1)=P(X = 0)=C(10,0)\times(0.25)^{0}\times(0.75)^{10}=0.0563$.

Part (c)

Answer:

$4 + \sqrt{3}$