7.5.1 Probability, Paper 1 (Short Questions)

Question 1:

The probability of student P being chosen as a school prefect is

\frac{3}{4}

$\frac{3}{4}$ while the probability of student Q being chosen is

\frac{5}{6}

$\frac{5}{6}$ .

Find the probability that

(a) both of the students are chosen as the school prefect,

(b) only one student is chosen as a school prefect.

Solution:

(a)

\begin{array}{l} Probability (both of the students are chosen as the school prefect) \\ = \frac{3}{4} \times \frac{5}{6} \\ = \frac{5}{8} \end{array}

$\begin{array}{l} Probability (both of the students are chosen as the school prefect) \\ = \frac{3}{4} \times \frac{5}{6} \\ = \frac{5}{8} \end{array}$

(b)

\begin{array}{l} Probability (only one student is chosen as a school prefect) \\ = (\frac{3}{4} \times \frac{1}{6}) + (\frac{1}{4} \times \frac{5}{6}) \\ = \frac{3}{24} + \frac{5}{24} \\ = \frac{1}{3} \end{array}

$\begin{array}{l} Probability (only one student is chosen as a school prefect) \\ = (\frac{3}{4} \times \frac{1}{6}) + (\frac{1}{4} \times \frac{5}{6}) \\ = \frac{3}{24} + \frac{5}{24} \\ = \frac{1}{3} \end{array}$

Question 2:

A bag contains x pink cards and 6 green cards. Two cards are drawn at random from the bag, one after the other, without replacement. Find the value of x if the probability of obtaining two green cards is ⅓.

Solution:

Total cards in the bag = x + 6

P(obtaining 2 green cards) = ⅓

\begin{array}{l} \frac{6}{x + 6} \times \frac{5}{x + 5} = \frac{1}{3} \\ \frac{30}{(x + 6) (x + 5)} = \frac{1}{3} \end{array}

$\begin{array}{l} \frac{6}{x + 6} \times \frac{5}{x + 5} = \frac{1}{3} \\ \frac{30}{(x + 6) (x + 5)} = \frac{1}{3} \end{array}$

(x + 6) (x + 5) = 90

x² + 11x + 30 = 90

x² + 11x – 60 = 0

(x – 4) (x + 15) = 0

x = 4 or x = –15 (not accepted)

Question 3:
A sample space of an experiment is given by S = {1, 2, 3, … , 21}. Events Q and R are defined as follows:
Q : {3, 6, 9, 12, 15, 18, 21}
R : {1, 3, 5, 15, 21}

Find
(a) P(Q)
(b) P(Q and R)

Solution:
(a)

\begin{array}{l} n (S) = 21, n (Q) = 7 \\ P (Q) = \frac{7}{21} = \frac{1}{3} \end{array}

$\begin{array}{l} n (S) = 21, n (Q) = 7 \\ P (Q) = \frac{7}{21} = \frac{1}{3} \end{array}$

(b)

\begin{array}{l} Q \cap R = {3, 15, 21}, \\ t h e n n (Q \cap R) = 3 \\ P (Q and R) = P (Q \cap R) \\ = \frac{n (Q \cap R)}{n (S)} \\ = \frac{3}{21} \\ = \frac{1}{7} \end{array}

$\begin{array}{l} Q \cap R = {3, 15, 21}, \\ t h e n n (Q \cap R) = 3 \\ P (Q and R) = P (Q \cap R) \\ = \frac{n (Q \cap R)}{n (S)} \\ = \frac{3}{21} \\ = \frac{1}{7} \end{array}$

Question 4:
The events A and B are not independent.

\begin{array}{l} Given P (A) = \frac{3}{5}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{1}{5}, find \\ (a) P [(A \cup B)'], \\ (b) P (A \cap B) . \end{array}

$\begin{array}{l} Given P (A) = \frac{3}{5}, P (B) = \frac{1}{4} and P (A \cup B) = \frac{1}{5}, find \\ (a) P [(A \cup B)'], \\ (b) P (A \cap B) . \end{array}$

Solution:
(a)

\begin{array}{l} P [(A \cup B)'] = 1 - P (A \cup B) \\ = 1 - \frac{1}{5} \\ = \frac{4}{5} \end{array}

$\begin{array}{l} P [(A \cup B)'] = 1 - P (A \cup B) \\ = 1 - \frac{1}{5} \\ = \frac{4}{5} \end{array}$

(b)

\begin{array}{l} P (A \cap B) = P (A) + P (B) - P (A \cup B) \\ = \frac{3}{5} + \frac{1}{4} - \frac{1}{5} \\ = \frac{13}{20} \end{array}

$\begin{array}{l} P (A \cap B) = P (A) + P (B) - P (A \cup B) \\ = \frac{3}{5} + \frac{1}{4} - \frac{1}{5} \\ = \frac{13}{20} \end{array}$

Leave a Comment Cancel reply