7.3 Probability of a Combined Event

Posted on April 24, 2020 by user

7.3 Probability of a Combined Event

7.3a Finding the Probability of a Combined Event by Listing the Outcomes

1. A combined event is an event resulting from the union or intersection of two or more events.

2. The union of combined event ‘A or B’ = A υ B

3. The intersection of combined event ‘A and B’ = A ∩ B

Example:

Diagram below shows five cards labelled with letters.

All these cards are put into a box. A two-letter code is to be formed by using any two of these cards. Two cards are picked at random, one after another, without replacement.

(a) List all sample space

(b) List all the outcomes of the events and find the probability that

(i) The code begins with the letter P.

(ii) The code consists of two vowel or two consonants.

Solution:

(a)

Sample space, S

= {(G, R), (G, A), (G, P), (G, E), (R, G), (R, A), (R, P), (R, E), (A, G), (A, R),

(A, P), (A, E), (P, G), (P, R), (P, A), (P, E), (E, G), (E, R), (E, A), (E, P)}

(b)

n(S) = 20

Let

A = Event of choosing a code begins with the letter P

B = Event of choosing the code consists of two vowel or two consonants.

(i)

A = {(P, G), (P, R), (P, A), (P, E)}

n(A) = 4

P (A) = \frac{4}{20} = \frac{1}{5}

(ii)

B = {(G, R), (G, P), (R, G), (R, P), (A, E), (P, G), (P, R), (E, A)}

n(B) = 8

P (B) = \frac{8}{20} = \frac{2}{5}

7.5 SPM Practice (Long Questions)

Posted on April 24, 2020 by user

Question 1:

All the cards written with the letters from the word ‘INTERNATIONAL’ are put into a box.

Two cards are drawn at random from the box, one after another, without replacement. Calculate the probability that

(a) the first card drawn has a letter N and the second card drawn has a letter I.

(b) the two cards drawn have the same letter.

Solution:
(a)
There are three cards with the letter ‘N’ and two card with letter ‘I’.
P (the first card drawn has a letter N and the second card drawn has a letter I)

\begin{array}{l} = \frac{3}{13} \times \frac{2}{(13 - 1)} \\ = \frac{3}{13} \times \frac{2}{12} = \frac{1}{26} \end{array}

(b)
P (the two cards drawn have the same letter)
= P (II or NN or TT or AA)
= P (II) + P (NN) + P (TT) + P (AA)

\begin{array}{l} = (\frac{2}{13} \times \frac{1}{12}) + (\frac{3}{13} \times \frac{2}{12}) + (\frac{2}{13} \times \frac{1}{12}) + (\frac{2}{13} \times \frac{1}{12}) \\ ↑ \\ \begin{array}{l} There are 3 letter ' N' \\ out of 13 letters \end{array} \\ = \frac{2}{156} + \frac{6}{156} + \frac{2}{156} + \frac{2}{156} \\ = \frac{12}{156} \\ = \frac{1}{13} \end{array}

Question 2:
During National Day celebration, a group of 8 boys and 5 girls from a school are taking part in a singing competition. Each day, two pupils are chosen at random to perform special skill.

(a) Calculate the probability that both pupils chosen to perform special skill are boys.

(b) Two boys were chosen to perform special skill on the first day. They are exempted from performing special skill on the second day.
Calculate the probability that both pupils chosen to perform special skill on the second day are of the same gender.

Solution:
(a)

\begin{array}{l} P (both pupils are boys) \\ = P (B B) \\ = \frac{8}{13} \times \frac{7}{12} \\ = \frac{7}{24} \end{array}

(b)

\begin{array}{l} P (both pupils are of the same gender) \\ = P (B B) + P (G G) \\ = (\frac{6}{11} \times \frac{5}{10}) + (\frac{5}{11} \times \frac{4}{10}) \leftarrow \begin{array}{l} 11 pupils left in the group to \\ perform special skill on the second \\ day after two boys were exempted . \end{array} \\ = \frac{3}{11} + \frac{2}{11} \\ = \frac{5}{11} \end{array}

7.1 Probability of an Event

Posted on April 24, 2020 by user

7.1 Probability of an Event

The probability of an event A, P(A) is given by

\begin{array}{l} P (A) = \frac{Number of times event A occurs}{Number of trials} \\ P (A) = \frac{n (A)}{n (S)} \\ where 0 \leq P (A) \leq 1 \end{array}

If P(A) = 0, then the event A will certainly not occur.

If P(A) = 1, then the event A is sure to occur.

Example 1:

A box contains 9 red pens and 13 blue pens. Tom puts another 4 red pens and 2 blue pens into the box. A pen is picked at random from the box. What is the probability that a red pen is picked?

Solution:

n(S) = 9 + 13 + 4 + 2 = 28

Let A = Event that a red pen is picked

n(A) = 9 + 4 = 13

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ = \frac{13}{28} \end{array}

Example 2:

A bag contains 45 green cards and yellow cards. A card is picked at random from the bag. The probability that a green card is picked is

\frac{1}{5} .

How many green cards must be added to the bag so that the probability of picking a green card becomes ½?

Solution:

n(S) = 45

Let

x = number of green cards in the bag.

A = Event of randomly picking a green card.

n(A) = x

\begin{array}{l} P (A) = \frac{n (A)}{n (S)} \\ \frac{1}{5} = \frac{x}{45} \\ x = \frac{45}{5} \\ x = 9 \end{array}

Let y is the number of green cards added to the bag.

\frac{9 + y}{45 + y} = \frac{1}{2}

2 (9 + y) = 45 + y

18 + 2y= 45 + y

2y – y = 45 – 18

y = 27