4 Marks Questions

Question 14 Marks

Two coins are tossed $400$ times and we get: Two heads: $112$ times; one head: $160$ times; $0$ head: $128$ times. When two coins are tossed at random, what is the probability of getting
$i. 2$ heads?
$ii. 1$ heads?
$iii. 0$ heads?

Answer

Total number of tosses $= 400$
Number of times $2$ heads appear $= 112$
Number of times $1$ head appears $= 160$
Number of times $0$ head appears $= 128$ In a random toss of two coins,
let $E_1, E_2, E_3$ be the events of getting $2$ heads,$1$ head and $0$ head, respectively. Then,
$i. P($getting $2$ heads) $= P(E_1)$ $=\frac{\text{Number of times $2$ heads appear}}{\text{Total number of trials}}$
$=\frac{112}{400}=0.28$
$ii. P($getting $1$ head) $= P(E_2)$ $=\frac{\text{Number of times $1$ head appear}}{\text{Total number of trials}}$
$=\frac{160}{400}=0.4$
$iii. P($getting $0$ head) $= P(E_3)$ $=\frac{\text{Number of times $0$ head appear}}{\text{Total number of trials}}$
$=\frac{128}{400}=0.32$
Remark: Clearly, when two coins are tossed, the only possible outcomes are $E_1, E_2$ and $E_3$ and $P(E_1) + P(E_2) + P(E_3) = (0.28 + 0.4 + 0.32) = 1$

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MCQ 24 Marks

A die is thrown $300$ times and the outcomes are noted as given below:

Outcome	$1$	$2$	$3$	$4$	$5$	$6$
Frequency	$60$	$72$	$54$	$42$	$39$	$33$

When a dice is thrown at random, what is the probability of getting a

A
$3?$
B
$6?$
C
$5?$
D
$1?$

Answer

Total number of tosses $=300 In$ a random throw of a dice, let $E_1, E_2, E_3, E_4$ be the events of getting $3,6,5$, and $1$ , respectively. Then,
i. $P($ getting $3$$)=P\left(E_1\right)=\frac{\text { Number of times } 3 \text { appear }}{\text { Total number of trials }}$
$=\frac{54}{300}=0.18$
ii. $P($ getting $6$$)=P\left(E_2\right)=\frac{\text { Number of times } 6 \text { appear }}{\text { Total number of trials }}$
$=\frac{33}{300}=0.11$
iii. $P($ getting $5$$)=P\left(E_3\right)=\frac{\text { Number of times } 5 \text { appear }}{\text { Total number of trials }}$
$=\frac{39}{300}=0.13$
iv. $P($ getting $1$$)=P\left(E_4\right)=\frac{\text { Number of times } 1 \text { appear }}{\text { Total number of trials }}$
$=\frac{60}{300}=0.20$

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MCQ 34 Marks

Three coins are tossed $200$ times and we get: Three heads: $39$ times; two heads: $58$ times; One head: $67$ times; $0$ head: $36$ times. When three coins are tossed at random, what is the probability of getting

A
$3$ heads?
B
$1$ heads?
C
$0$ heads?
D
$2$ heads?

Answer

Total number of tosses $=200$ Number of times $3$ heads appear $=39$ Number of times $2$ head appears $=58$ Number of times $1$ head appears $=67$ Number of times $0$ head appears $=36$ In a random toss of three coins, let $E _1, E _2, E _3, E _4$ be the events of getting $3$ heads, $2$ heads, $1$ head and $0$ head, respectively. Then,
$i. P($getting $3$ heads) $= P(E_1)$ $=\frac{\text{Number of times $3$ heads appear}}{\text{Total number of trials}}$
$=\frac{39}{200}=0.195$
$ii. P($getting $1$ heads) $= P(E_2)$ $=\frac{\text{Number of times $1$ head appear}}{\text{Total number of trials}}$
$=\frac{67}{200}=0.335$
$iii. P($getting $0$ head) $= P(E_3)$ $=\frac{\text{Number of times $0$ head appear}}{\text{Total number of trials}}$
$=\frac{36}{200}=0.18$
$iv. P($getting $2$ heads) $= P(E_4)$ $=\frac{\text{Number of times $2$ heads appear}}{\text{Total number of trials}}$
$=\frac{58}{200}=0.29$
Remark: Clearly, when two coins are tossed, the only possible outcomes are $E_1, E_2, E_3$ and $E_4$ and $P\left(E_1\right)+P\left(E_2\right)+$ $P\left(E_3\right)+P\left(E_4\right)=(0.195+0.335+0.18+0.29)=1$

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