ΙΒ μαθηματικά HL- Combinatorics – Probability

Permutations Combinations – IB Mathematics HL

  1. Find the number of possible choices for choosing a sub-committee of five persons consisting of the chairperson, secretary and three ordinary membersfrom a committee of 40 people.

Solution

The chairperson can be chosen by\(^40\mathrm{C}_1\) ways

The secretary can be chosen by \(^39\mathrm{C}_1\) ways

And the three ordinary members can be chosen by \(^38\mathrm{C}_3\) ways

So, the members of this sub-committee can be chosen by

\(^40\mathrm{C}_1 \cdot ^39\mathrm{C}_1 \cdot ^38\mathrm{C}_3=\)

\(=40 \cdot 39\cdot8436=13,160,160\) ways.

 

2. In a class of 12 students, 4 are male and 8 are female. How can we find the number of committees of 5 members can be formed containing 2 males and 3 females?

Solution

The males can be chosen by \(^4\mathrm{C}_2\) ways

The females, can be chosen by \(^8\mathrm{C}_3\) ways

So, the members of this committee can be chosen by

\(^4\mathrm{C}_2 \cdot ^8\mathrm{C}_3=\)

\(=6 \cdot 56=336\) ways.

 

3. Find the number of different arrangements of the letters of the word “LONDON”? and how can we calculate the probability of the event that the middle two letters both are N’s.

Solution

The number of different arrangements of the six letters from which there are two pairs of same letters is given by the following formula

\(\frac{6!}{2!2!}=\frac{720}{4}=180\)

Since the two N’s are located in the middle of the word there are \(\frac{4!}{2!}=\frac{24}{2}=12\) ways to arrange the remaining 4 letters from which there are two O,s.

So, the probability equals to \(\frac{12}{180}=\frac{1}{15} \)

 

4. Find the probability \(P(A\cap B’)\) if A and B are two events such that \(P(A)=0.5\), \(P(B)=0.3\) and \(P(A \cup B)=0.7\).

 

Solution

We know that

\(P(A \cup B)=P(A)+P(B)-P(A \cap B) \Rightarrow \) \(P(A \cap B)=0.5+0.3-0.7=0.1 \)

Finally, we have

\(P(A \cap B’)=P(A)-P(A \cap B) =0.5-0.1=0.4 \)

 

Conditional Probability – IB Mathematics HL – IB maths HL

5. Find the probability \(P(A/B’)\) if A and B are two events such that \(P(A)=0.4\), \(P(B)=0.2\) and \(P(A/B)=0.8\).

Solution

From conditional probability formula we have that

\(P(A/B)=\frac{P(A \cap B)}{P(B)}\Rightarrow \) \(0.8=\frac{P(A \cap B)}{0.2}\Rightarrow P(A \cap B) =0.8 \cdot 0.2=0.16\)

Finally, we have that

\(P(A/B’)=\frac{P(A \cap B’)}{P(B’)}=\frac{P(A)- P(A \cap B)}{1-P(B)}=\) \(=\frac{0.4-0.16}{1-0.2}=\frac{0.24}{0.8}=0.3 \)