5 Marks Questions

Question 15 Marks

Determine the probability p, for each of the following events.

An odd number appears in a single toss of a fair die.
At least one head appears in two tosses of a fair coin.
A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.
The sum of 6 appears in a single toss of a pair of fair dice.

Answer

When a die is thrown the possible outcomes are

S = {1, 2, 3,4, 5, 6} out of which 1, 3, 5 are odd,

$\therefore\ \text{Required probability}=\frac{3}{6}=\frac{1}{2}$

When a fair coin is tossed two times the space is

S = {HH, HT, TH, TT}

If at least one head appears then the favourable casses are HH, HT and TH.

$\therefore\ \text{Required probability}=\frac{3}{4}$

When a pair of dice is rolled, total number of cases = 6 × 6 = 36

If sum is 6 then possible outcomes are (1, 5), (5, 1), (2, 4), (4, 2) and (3, 3)

$\therefore\ \text{Required probability}=\frac{5}{36}$

(missing)

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Question 25 Marks

Four candidates A, B, C, D have applied for the assignment to coach a school cricket team. If A is twice as likely to be selected as B, and B and C are given about the same chance of being selected, while C is twice as likely to be selected as D, what are the probabilities that:

C will be selected?
A will not be selected?

Answer

Given that A is twice as likely to be selected as B
I.e., $\text{P(A)}=2\text{P(B)}$
and C is twice as likely to be selected as D
$\therefore\ \text{P(C)}=2\text{P(D)}$
$\Rightarrow\text{P(B)}=2\text{P(D)}$
$\Rightarrow\frac{\text{P(A)}}{2}=2\text{P(D)}$
$\Rightarrow\text{P(D)}\frac{1}{4}\text{P(A)}$
Now B and C are given about the same chance
$\therefore\ \text{P(B)}=\text{P(C)}$
Since, sum of all probabilities = 1
$\therefore\ \text{P(A)}+\text{P(B)}+\text{P(C)}+\text{P(D)}=1$
$\Rightarrow\text{P(A)}+\frac{\text{P(A)}}{2}+\frac{\text{P(A)}}{2}+\frac{\text{P(A)}}{4}=1$
$\Rightarrow4\text{P(A)}+2\text{P(A)}+2\text{P(A)}+\text{P(A)}=4$
$\Rightarrow9\text{P(A)}=4$
$\Rightarrow\text{P(A)}=\frac{4}{9}$

P(C will be selected)

$=\text{P(C)}=\text{P(B)}=\frac{\text{P(A)}}{2}$
$=\frac{4}{9}\times\frac{1}{2}=\frac{2}{9}$

P(A will not be selected)

$\text{P}(\text{A}')=1-\text{P(A)}$
$=1-\frac{4}{9}=\frac{5}{9}$

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Question 35 Marks

One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted} You are told that the chances of John’s promotion is same as that of Gurpreet, Rita’s chances of promotion are twice as likely as Johns. Aslam’s chances are four times that of John:

Determine P(John promoted)

P(Rita promoted)
P(Aslam promoted)
P(Gurpreet promoted)

If A = {John promoted or Gurpreet promoted}, find P(A).

Answer

Let Event, J = John promoted
R = Rita promoted
A = Aslam promoted
G = Gurpreet promoted
Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}
i.e. S = {J, R, A, G}
t is given that, chances of John’s promotion is same as that of Gurpreet.
P(J) = P(G)
Rita’s chances of promotion are twice as likely as John.
P(R) = 2P(J)
And Aslam’s chances of promotion are four times that of John.
P(A) = 4P(J)
Now, P(J) + P(R) + P(A) + P(G) = 1
⇒ P(J) + 2P(J) + 4P(J) + P(J) = 1
⇒ P(J) = 1
$\Rightarrow\text{P(J)}=\frac{1}{8}$

P(John promoted)

$=\text{P(J)}=\frac{1}{8}$
P(Rita promoted)
$=\text{P(R)}=2\text{P(J)}=2\times\frac{1}{8}=\frac{1}{4}$
A(Aslam promoted)
$=\text{P(A)}=4\text{P(J)}=4\times\frac{1}{8}=\frac{1}{2}$
P(Gurpreet promoted)
$=\text{P(G)}=\text{P(J)}=\frac{1}{8}$

A = John promoted or Gurpreet promoted

$\therefore\ \text{A}=\text{J}\cup\text{G}$
$\text{P(A)}=\text{P(J}\cup\text{G})$
$=\text{P(J)}+\text{P(G)}-\text{P}(\text{J}\cap\text{G})$
$=\frac{1}{8}+\frac{1}{8}-0$ $\big[\because\text{P(J}\cap\text{G})=0\big]$
$=\frac{1}{4}$

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Question 45 Marks

The accompanying Venn diagram shows three events, A, B, and C, and also the probabilities of the various intersections $($for instance, $\text{P}(\text{A}\cap\text{B})=.07)$ Determine:

$\text{P(A)}$
$\text{P}(\text{B}\cap\bar{\text{C}})$
$\text{P(A}\cup\text{B})$
$\text{P(A}\cap\bar{\text{B}})$
$\text{P(B}\cap\text{C})$
Probability of exactly one of the three occurs.

Answer

From the given Venn diagram.

$\text{P(A)}=0.13+0.07=0.20$
$\text{P}(\text{B}\cap\bar{\text{C}})=\text{P(B)}-\text{P}(\text{B}\cap\text{C})$

$=0.07+0.10+0.15-0.15$
$=0.07+0.10$
$=0.17$

$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$

$=0.13+0.07+0.07+0.10+0.15-0.07$
$=0.13+0.07+0.10+0.15$
$=0.45$

$\text{P}(\text{A}\cap\bar{\text{B}})=\text{P(A)}-\text{P}(\text{A}\cap\text{B})$

$=0.13+0.07-0.07$
$=0.03$

$\text{P(B}\cap\text{C})=0.15$
P(exactly one of the three occurs)

$=0.13+0.10+0.28$
$=0.51$

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Question 55 Marks

Match the following:

a.	If E₁ and E₂ are the two mutually exclusive events	i.	$\text{E}_1\cap\text{E}_2=\text{E}_1$
b.	If E₁ and E₂ are the mutually exclusive and exhaustive events	ii.	$(\text{E}_1-\text{E}_2)\cup(\text{E}_1\cap\text{E}_2)=\text{E}_1$
c.	If E₁ and E₂ have common outcomes, then	iii.	$\text{E}_1\cap\text{E}_2=\phi,\text{ E}_1\cup\text{E}_2=\text{S}$
d.	If E₁ and E₂ are two events such that $\text{E}_1\subset\text{E}_2$	iv.	$\text{E}_1\cap\text{E}_2=\phi$

Answer

Match the following:

a.	If E₁ and E₂ are the two mutually exclusive events	iv.	$\text{E}_1\cap\text{E}_2=\phi$
b.	If E₁ and E₂ are the mutually exclusive and exhaustive events	iii.	$\text{E}_1\cap\text{E}_2=\phi,\text{ E}_1\cup\text{E}_2=\text{S}$
c.	If E₁ and E₂ have common outcomes, then	ii.	$(\text{E}_1-\text{E}_2)\cup(\text{E}_1\cap\text{E}_2)=\text{E}_1$
d.	If E₁ and E₂ are two events such that $\text{E}_1\subset\text{E}_2$	i.	$\text{E}_1\cap\text{E}_2=\text{E}_1$

Solution:

If E₁ and E₂ are mutually exclusive events, then $\text{E}_1\cap\text{E}_2=\phi$
If E₁ and E₂ are the mutually exclusive and exhaustive events $\text{E}_1\cap\text{E}_2=\phi,$ and $\text{E}_1\cup\text{E}_2=\text{S}$
If E₁ and E₂ have common outcomes, then $(\text{E}_1-\text{E}_2)\cup(\text{E}_1\cap\text{E}_2)=\text{E}_1$

If E₁ and E₂ are two events such that

$\text{E}_1\subset\text{E}_2\Rightarrow\text{E}_1\cap\text{E}_2=\text{E}_1$

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Question 65 Marks

Match the proposed probability under Column C₁ with the appropriate written description under column C₂:

	C₁		C₂
	Probability		Written Description.
a.	0.95	i.	An incorrect assignment.
b.	0.02	ii.	No chance of happening.
c.	-0.3	iii.	As much chance of happening as not.
d.	0.5	iv.	Very likely to happen.
e.	0	v.	Very little chance of happening.

Answer

	C₁		C₂
	Probability		Written Description.
a.	0.95	iv.	Very likely to happen.
b.	0.02	v.	Very little chance of happening.
c.	-0.3	i.	An incorrect assignment.
d.	0.5	iii.	As much chance of happening as not.
e.	0	ii.	No chance of happening.

Solution:

0.95 = Very likely to happen, so it is close to 1.
0.02 = Very little chance of happening as the probability is very low.
-0.3 = an incorrect assignment because probability is never negative.
0.5 = As much chance of happening as not because sum of chances of happening and not happening is one.
0 = No chance of happening.

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Question 75 Marks

A team of medical students doing their internship have to assist during surgeries at a city hospital. The probabilities of surgeries rated as very complex, complex, routine, simple or very simple are respectively, 0.15, 0.20, 0.31, 0.26, 08. Find the probabilities that a particular surgery will be rated:

Complex or very complex.
Neither very complex nor very simple.
Routine or complex.
Routine or simple.

Answer

Let E₁, E₂, E₃, E₄, and E₅be the event that surgeries are rated as very complex, complex, rautine, simple or very simple, respectively.
$\therefore\ \text{P}(\text{E}_1)=0.15,\text{ P}(\text{E}_2)=0.20,\text{ P}(\text{E}_3)=0.31,\text{ P}(\text{E}_4)=0.26,\text{ P}(\text{E}_5)=0.08$

P(complex or very complex),

$=\text{P}(\text{E}_1\text{ or E}_2)=\text{P}(\text{E}_1\cup\text{E}_2)$
$=\text{P}(\text{E}_1)+\text{P}(\text{E}_2)-\text{P}(\text{E}_1\cap\text{E}_2)$
$=0.15+0.20-0$
$=0.35$

P(neither very complex nor very simple),

$\text{P}(\text{E}'_1\cap\text{E}'_5)=\text{P}(\text{E}_1\cup\text{E}_5)'$
$=1-\text{P}(\text{E}_1\cup\text{E}_5)$
$=1-\big[\text{P}(\text{E}_1)+\text{P}(\text{E}_5)\big]$
$=1-(0.15+0.08)$
$=1-0.23$
$=0.77$

P(routine or complex),

$=\text{P}(\text{E}_3\cup\text{E}_2)=\text{P}(\text{E}_3)+\text{P}(\text{E}_2)$
$=0.31+0.20$
$=0.57$

P(routine or simple),

$=\text{P}(\text{E}_3\cup\text{E}_4)=\text{P}(\text{E}_3)+\text{P}(\text{E}_4)$
$=0.31+0.26$
$=0.57$

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Question 85 Marks

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. Find P(G), where G is the event that a number greater than 3 occurs on a single roll of the die.

Answer

Given that probability of even numbers,
$=\frac{1}{2}\times\text{Probability of odd numbers}$
$\Rightarrow\text{P(Odd)}:\text{P(Even)}=2:1$
$\therefore\ \text{P(Odd number)}=\frac{2}{2+1}=\frac{2}{3}$
and $\text{P(even number)}=\frac{1}{2+1}=\frac{1}{3}$
Also given that, G the event that a number greater 3 occurs in single throw of die.
$\therefore$ The possible outcome are 4, 5 and 6 out of which two are even and one is odd.
$\therefore\ \text{Required probability}=\text{P(G)}$
$=2\times\text{P(even)}\times\text{P(odd)}$
$=2\times\frac{1}{3}\times\frac{2}{3}=\frac{4}{9}$
Hence, the required probability is $\frac{4}{9}$

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Question 95 Marks

In a large metropolitan area, the probabilities are 87, 36, 30 that a family (randomly chosen for a sample survey) owns a colour television set, a black and white television set, or both kinds of sets. What is the probability that a family owns either anyone or both kinds of sets?

Answer

Let E₁ be the even that a family owns colour television set E₂ be the even that the family.
owns black and white television set.
Given that P(E₁) = 0.87, P(E₂) = 0.36
and $\text{P}(\text{E}_1\cap\text{E}_2)=0.30$
$\therefore$ The probability that a family owns either colour television set or black and white television set.
$\therefore\ \text{P}(\text{E}_1\cup\text{E}_2)=\text{P(E)}_1+\text{P(E)}_2-\text{P}(\text{E}_1\cap\text{E}_2)$
$=0.87+0.36-0.30$
$=0.93$
Hence, the required probability = 0.93

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Question 105 Marks

A sample space consists of 9 elementary outcomes e₁, e₂, ...., e₉ whose probabilities are
P(e₁) = P(e₂ ) = 0.08, P(e₃ ) = P(e₄) = P(e₅) = 0.1
P(e₆) = P(e₇) = 0.2, P(e₈) = P(e₉) = 0.07
Suppose A = {e₁, e₅, e₈}, B = {e₂, e₅, e₈, e₉}

Calculate P (A), P (B), and $\text{P}(\text{A}\cap\text{B})$
Using the addition law of probability, calculate $\text{P}(\text{A}\cup\text{B})$
List the composition of the event $\text{A}\cup\text{B},$ and calculate $\text{P}(\text{A}\cup\text{B})$ by adding the probabilities of the elementary outcomes.
Calculate $\text{P}(\bar{\text{B}})$ from P(B), also calculate $\text{P}(\bar{\text{B}})$ directly from the elementary outcomes of $\bar{\text{B}}$

Answer

$\text{P(A)}=\text{P}(\text{E}_1)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)$

$\text{P(A)}=0.08+0.1+0.07=0.25$

$\text{P(B)}=\text{P}(\text{E}_2)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)+\text{P}(\text{E}_9)$

$\text{P(B)}=0.08+0.1+0.07+0.07=0.32$

$\text{P}(\text{A}\cup\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$

Now, $\text{A}\cap\text{B}=\big\{\text{E}_5,\text{E}_8\big\}$

$\therefore\ \text{P}(\text{A}\cap\text{B})=\text{P}(\text{E}_5)+\text{P}(\text{E}_8)$

$=0.1+0.07=0.17$

$\therefore\ \text{P}(\text{A}\cup\text{B})=0.25+0.32-0.17=0.40$

$\text{A}\cup\text{B}=\big\{\text{E}_1,\text{E}_2,\text{E}_5,\text{E}_8,\text{E}_9\big\}$

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{E}_1)+\text{P}(\text{E}_2)+\text{P}(\text{E}_5)+\text{P}(\text{E}_8)+\text{P}(\text{E}_9)$

$=0.08+0.08+0.1+0.07+0.07=0.40$

$\because\ \text{P}\bar{(\text{B})}=1-\text{P(B)}$

$=1-0.32=0.68$

and $\bar{\text{B}}=\big\{\text{E}_1,\text{E}_3,\text{E}_4,\text{E}_6,\text{E}_7\big\}$

$\therefore\ \text{P}(\bar{\text{B}})=\text{P}(\text{E}_1)+\text{P}(\text{E}_3)+\text{P}(\text{E}_4)+\text{P}(\text{E}_6)+\text{P}(\text{E}_7)$

$=0.08+0.1+0.1+0.2+0.2=0.68$

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MATHS 5 Marks Questions

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