5 Marks Questions

Question 15 Marks

$1500$ families with $2$ children were selected randomly, and the following data were recorded:

No of girls in a family	$0$	$1$	$2$
No of girls	$211$	$814$	$475$

If a family is chosen at random, compute the probability that it has:
$1.$ No girl.
$2. 1$ girl.
$3. 2$ girls.
$4.$ At most one girl.
$5.$ More girls than boys.

Answer

$1.$ Probability of having no girl in a family $=\frac{\text{No. of families having no girl}}{\text{Total no. of families}}$
$=\frac{211}{1500}=0.1406$
$2.$ Probability of having $1$ girl in a family $=\frac{\text{No. of families having 1 girl}}{\text{Total no. of families}}$
$=\frac{814}{1500}$
$=\frac{407}{750}=0.5426$
$3.$ Probability of having $2$ girl in a family $=\frac{\text{No. of families having 2 girl}}{\text{Total no. of families}}$
$=\frac{475}{1500}=0.3166$
$4.$ Probability of having at the most one girl $=\frac{\text{No. of families having at most one girl}}{\text{Total no. of families}}$
$=\frac{211+814}{1500}$
$=\frac{1025}{1500}=0.6833$
$5.$ Probability of having more girls than boys $=\frac{\text{No. of families having more girls than boys}}{\text{Total no. of families}}$
$=\frac{475}{1500}=0.31$

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Question 25 Marks

Three coins are tossed simultaneously $100$ times with the following frequencies of different outcomes:

Outcome	No head	One head	Two head	Three head
Frequency	$14$	$38$	$36$	$12$

If the three coins are tossed simultaneously again, compute the probability of:
$1.$ heads coming up.
$2.$ heads coming up.
$3.$ At least one Head coming up.
$4.$ Getting more Tails than Heads.
$5.$ Getting more heads than tails.

Answer

$1.$ Probability of $2$ heads coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{36}{100}=0.36$
$2.$ Probability of $3$ heads coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{12}{100}=0.12$
$3.$ Probability of at least one head coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{38+36+12}{100}$
$=\frac{86}{100}=0.86$
$4.$ Probability of getting more heads than Tails $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{36+12}{100}$
$=\frac{48}{100}=0.48$
$5.$ Probability of getting more tails than heads $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{14+38}{100}$
$=\frac{52}{100}=0.52$

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Question 35 Marks

The following table gives the life time of $400$ neon lamps:

Life time$($in hours$)$	$300 - 400$	$400 - 500$	$500 - 600$	$600 - 700$	$700 - 800$	$800 - 900$	$900 -1000$
Number of lamps:	$14$	$56$	$60$	$86$	$74$	$62$	$48$

A bulb is selected at random. Find the probability that the lifetime of a selected bulb is:
$1.$ Less than $400$
$2.$ between $300$ to $800$ hours
$3$. At least $700$ hours

Answer

Total number of bulbs $= 400$
$1.$ Probability that the life of the selected bulb is less than $400$ hrs
$=\frac{\text{No. of bulbs having life less than 400 hrs}}{\text{Total no. of bulbs}}$
$=\frac{14}{400}$
$=\frac{7}{200}$
$2.$ Probability that the life of the selected bulb is between $300 - 800$ hrs
$=\frac{\text{No. of bulbs having life less than 400 hrs}}{\text{Total no. of bulbs}}$
$=\frac{14+56+60+86+74}{200}$
$=\frac{290}{400}$
$=\frac{29}{40}$
$3.$ Probability that the life of the selected bulb is at least $700$ hrs
$=\frac{\text{No. of bulbs having life less than 700 hrs}}{\text{Total no. of bulbs}}$
$=\frac{74+62+48}{400}$
$=\frac{184}{400}$
$=\frac{23}{50}$

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Question 45 Marks

Define probability of an event.

Answer

The probability of an event denotes the relative frequency of occurrence of an experiment’s outcome, when repeating the experiment.
Definition:
The empirical or experimental definition of probability is that if $n$ be the total number of trials of an experiment and $A$ is an event associated to it such that $A$ happens in $m-$trials, then the probability of happening of event $A$ is denoted by $P(A)$ and is given by $\text{P(A)}=\frac{\text{m}}{\text{n}}$ To illustrate the definition, let us take examples:
$1.$ When two coins are tossed simultaneously, the possible outcomes are $HH, HT, TH$ and $TT.$
The total number of trials is $4.$
$2.$ Let $A$ be the event of occurring exactly two heads.
$3.$ The number of times $A$ happens is $1.$
$4.$ So, the probability of the event $A$ is
$\text{P(A)}=\frac{\text{m}}{\text{n}}$
$=\frac{1}{4}$
$=0.25$
$1.$ In the experiment of rolling a dice, the possible outcomes are $1, 2, 3, 4, 5$ and $6.$
$2.$ Let $A$ be the event of occurring a number greater than $3.$
$3.$ The total number of trials is $6.$ The number of times $A$ happens is $3.$
$4.$ So, the probability of the event $A$ is
$\text{P(A)}=\frac{\text{m}}{\text{n}}$
$=\frac{3}{6}$
$=\frac{1}{2}$
$=0.5$
Note that $H$ stands for getting a head and $T$ stands for getting a tail in the experiment of tossing a coin.

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