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

$2$ heads?
$1$ heads?
$0$ 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 $0$ hvead, respectively. Then,

$P($getting $2$ heads$) = P(E_1) =\frac{\text{Number of times 2 heads appear}}{\text{Total number of trials}}$

$=\frac{112}{400}=0.28$

$P($getting $1$ head$) = P(E_2) =\frac{\text{Number of times 1 head appear}}{\text{Total number of trials}}$

$=\frac{160}{400}=0.4$

$P($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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Question 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

$3$?
$6$?
$5$?
$1$?

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,

$P($getting $3) = P(E_1) =\frac{\text{Number of times 3 appear}}{\text{Total number of trials}}$

$=\frac{54}{300}=0.18$

$P($getting $6) = P(E_2) =\frac{\text{Number of times 6 appear}}{\text{Total number of trials}}$

$=\frac{33}{300}=0.11$

$P($getting $5) = P(E_3) =\frac{\text{Number of times 5 appear}}{\text{Total number of trials}}$

$=\frac{39}{300}=0.13$

$P($getting $1) = P(E_4) =\frac{\text{Number of times 1 appear}}{\text{Total number of trials}}$

$=\frac{60}{300}=0.20$

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Question 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

$3$ heads?
$1$ heads?
$0$ heads?
$2$ 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,

$P($getting $3$ heads$) = P(E_1) =\frac{\text{Number of times 3 heads appear}}{\text{Total number of trials}}$

$=\frac{39}{200}=0.195$

$P($getting $1$ heads$) = P(E_2) =\frac{\text{Number of times 1 head appear}}{\text{Total number of trials}}$

$=\frac{67}{200}=0.335$

$P($getting $0$ head$) = P(E_3) =\frac{\text{Number of times 0 head appear}}{\text{Total number of trials}}$

$=\frac{36}{200}=0.18$

$P($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 $P(E_1) + P(E_2) + P(E_3) + P(E_4) = (0.195 + 0.335 + 0.18 + 0.29) = 1$

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