5 Marks Questions

Question 15 Marks

$A$ and $B$ are two events such that $P(A) = 0.54, P(B) = 0.69$ and $P(A \cap B) = 0.35$. Find

$P(A \cup B)$
$P(A'\cap B')$
$P(A\cap B')$
$P( B\cap A')$

Answer

Here $P(A) = 0.54, P(B) = 0.69$ and $\style{font-size:36px}{\mathrm P(\mathrm A\cap\mathrm B)} = 0.35$

We know that
$P ( A \cup B ) = P ( A ) + P ( B ) - P ( A \cap B )$
$= 0.54+ 0.69 - 0.35$
$= 1.23 - 0.35 = 0.88$
$\style{font-size:36px}{\mathrm P(\mathrm A'\cap\mathrm B')=\mathrm P(\mathrm A\cup\mathrm B)'=1-\mathrm P(\mathrm A\cup\mathrm B)}$
$= 1 - 0.88 = 0.12$
$\style{font-size:36px}{\mathrm P(\mathrm A\cap\mathrm B')=\mathrm P(\mathrm A)-\mathrm P(\mathrm A\cap\mathrm B)}$
$= 0.54 - 0.35 = 0.19$
$P(B\cap A') = P(B)- P(A\cap B)$
$= 0.69 - 0.35 = 0.34$

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Question 25 Marks

If $\frac2{11}$ is the probability of an event A, what is the probability of the event 'not A'.

Answer

Given, probability of event A i.e., P(A) = $\frac2{11}$
$\therefore$ Probability ('not A')= $P(\bar A)$
$=1-P(A)=1-\frac2{11}=\frac9{11}$

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Question 35 Marks

In a lottery, a person chosen, six different natural numbers at random from 1 to 20 and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game?
[Hint order of the numbers is not important.]

Answer

Number of numbers in the draw = 20
Number of numbers to be selected = 6
Let A be the event that six numbers match with the six numbers fixed by the lottery committee.
$\therefore \;n(A){ = ^6}{C_6} = 1$
Thus probability of winning the prize P(A) = $ \frac{{n(A)}}{{n(S)}} = \frac{{^6{C_6}}}{{^{20}{C_6}}}$
$= \frac{1}{{20!}} \times 6! \times 14! = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 14!}}{{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14!}}$
$= \frac{1}{{38760}}$

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Question 45 Marks

An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:
A: the sum is greater than 8, B: 2 occurs on either die.
C: the sum is at least 7 and a multiple of 3
which pairs of these events are mutually exclusive?

Answer

When a pair of die is rolled then the sample space (S) is given by,
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A : the sum is greater than 8.

$\therefore$ A = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
B : 2 occurs on either die :

$\therefore$ B = {(1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (2,1), (2, 3), (2, 4), (2,5) (2, 6)}
C: the sum is at least 7 and a multiple of 3.
$\therefore$ C = {(3, 6), (6, 3), (5, 4), (4, 5), (6, 6)}
Now $\style{font-size:28px}{A\cap B=\phi}$
$\therefore$ A and B are mutually exclusive events.

$\style{font-size:28px}{B\cap C=\phi}$
B and C are mutually exclusive events.
$\style{font-size:28px}{A\cap C}$= {(3, 6), (4, 5), (5, 4), (6, 3), (6, 6)} $\style{font-size:28px}{⧧\phi}$
Thus A and C are not mutually exclusive events.

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Question 55 Marks

A die is thrown. Describe the following events:

$A:$ a number less than $7$
$B:$ a number greater than $7$
$C:$ a multiple of $3$
$D:$ a number less than $4$
$E:$ an even number greater than $4$
$F:$ a number not less than $3$

Also find $\ce{{A\cup B}, {A\cap B}, {B\cup C}, {E\cap F}, {D\cap E}, A –C, D–E, {E\cap F'}, F'.}$

Answer

When a die is thrown then, the sample space is given by, $S = \{1, 2, 3, 4, 5, 6\}$

$A:$ a number less than $7 = \{1, 2, 3, 4, 5, 6\}$
$B:$ a number greater than $7 = \phi$
$C:$ a multiple of $3 = \{3, 6\}$
$D:$ a number less than, $4 = \{1, 2, 3\}$
$E:$ an even number greater than $4 = \{6\}$
$F:$ a number not less than $3 = \{3, 4, 5, 6\}$

Now : $A\cup B=\{1,\;2,\;3,\;4,\;5,\;6\}$,
$A \cap B = \phi$,
$B\cup C=\{3,6\}$,
$E\cap F=\{6\}$,
$D \cap E = \phi$,
$A - C = \{1, 2, 4, 5\},$
$D - E = \{1, 2, 3\},$
Since, $F'=\{1,\;2\}$
$\therefore $ $E\cap F'=\{\phi\}$,
$F'=\{1,\;2\}$

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Question 65 Marks

A bag contains $9$ discs of which $4$ are red, $3$ are blue and $2$ are yellow. The discs are similar in shape and size. A disc is drawn at random from the bag. Calculate the probability that it will be

red
yellow
blue
not blue
either red or blue.

Answer

There are $9$ discs in all so the total number of possible outcomes $= n(S) = 9.$
Let the events $A, B, C$ be defined as
$A:$ ‘the disc drawn is red’
$B:$ ‘the disc drawn is yellow’
$C:$ ‘the disc drawn is blue’.So, now

The number of red discs $=$ number of favourable outcomes$= 4$, i.e., $n (A) = 4$
Hence $P(A) = \frac 49$
The number of yellow discs $=$ number of favourable outcomes$= 2$, i.e., $n (B) = 2$
Therefore, $P(B) = \frac 29$
The number of blue discs $=$ number of favourable outcomes$= 3,$ i.e., $n(C) = 3$
Therefore, $P(C) = \frac 39 = \frac 13$
Clearly the event ‘not blue’ is ‘not $C’$. We know that $P($not $C) = 1 – P(C)$
Therefore $P($not $C) = 1-\frac 13 = \frac 23$
The event ‘either red or blue’ may be described by the set $‘A$ or $C’$
Since, $A$ and $C$ are mutually exclusive events, we have
$P(A$ or $\ce{C) = P (A \cup C) = P(A) + P(C)}=\frac 43 + \frac 13 = \frac 79$

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