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Maths 5 Mark Question

Probability · Maharashtra Board STD 9 (MSBSHSE - English Medium). 4 questions.

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5 Mark Question

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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 $=\frac{\text{No. of families having no girl}}{\text{Total no. of families}}$
$=\frac{211}{1500}=0.1406$
  1. Probability of having 1 girl in a family $=\frac{\text{No. of families having 1 girl}}{\text{Total no. of families}}$
$=\frac{814}{1500}$

$=\frac{407}{750}=0.5426$
  1. Probability of having 2 girl in a family $=\frac{\text{No. of families having 2 girl}}{\text{Total no. of families}}$
$=\frac{475}{1500}=0.3166$
  1. Probability of having at the most one girl $=\frac{\text{No. of families having at most one girl}}{\text{Total no. of families}}$
$=\frac{211+814}{1500}$

$=\frac{1025}{1500}=0.6833$
  1. 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. 2 heads coming up.
  2. 3 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 $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{36}{100}=0.36$
  1. Probability of 3 heads coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{12}{100}=0.12$
  1. Probability of at least one head coming up $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{38+36+12}{100}$

$=\frac{86}{100}=0.86$
  1. Probability of getting more heads than Tails $=\frac{\text{Favorable out come}}{\text{Total out come}}$
$=\frac{36+12}{100}$

$=\frac{48}{100}=0.48$
  1. 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
$=\frac{\text{No. of bulbs having life less than 400 hrs}}{\text{Total no. of bulbs}}$

$=\frac{14}{400}$

$=\frac{7}{200}$
  1. Probability that the life of the selected bulb is between 300 - 800 hrs
$=\frac{\text{No. of bulbs having life less than 400 hrs}}{\text{Total no. of bulbs}}$

$=\frac{14+56+60+86+74}{200}$

$=\frac{290}{400}$

$=\frac{29}{40}$
  1. 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. Let A be the event of occurring exactly two heads. The number of times A happens is 1. 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. Let A be the event of occurring a number greater than 3. The total number of trials is 6. The number of times A happens is 3. 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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