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Maths 3 Marks Question

Probability · Jharkhand Board (JAC) STD 9 (Jharkhand - English Medium). 7 questions.

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Question 13 Marks
In a circket match, a batsman hits a boundary $6$ times out of $30$ balls he plays. Find the probability that he did not hit a boundary.
Answer
Total number of ball played $= 30$
Number of times boundary was hit $= 6$
$\Rightarrow$ Number of times boundary was not hit $= 30 - 6 = 24$
Therefore, Probability that the batsman did not hit the boundary $=\frac{\text{No. of times boundary was not hit}}{\text{Total number of balls played}}$
$=\frac{24}{30}$
$=\frac{4}{5}=0.8$
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Question 23 Marks
On one page of a telephone directory, there are $200$ phone numbers. The frequency distribution of their unit's digits is given below:
Unit's digit $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$
Frequency $19$ $22$ $23$ $19$ $21$ $24$ $23$ $18$ $16$ $15$
One of the numbers is chosen at random from the page. What is the probability that the unit's digit of the chosen number is:
$i. 5?$
$ii. 8?$
Answer
Total phone numbers on the directory page $= 200$
$i.$ Number of numbers with units digits $5 = 24$
Let $E_1$ be the event that the units digits of selected number is $5.$
$\therefore$ Required probability $= P(E_1) =\frac{24}{200}=0.12$
$ii.$ Number of numbers with units digits $8 = 16$
Let $E_2$ be the event that the units digits of selected number is $8.$
$\therefore$ Required probability $= P(E_2) =\frac{16}{200}=0.08$
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Question 33 Marks
The following table shows the blood groups of $40$ students of a class.
Blood group $A$ $B$ $O$ $AB$
Number of students $11$ $9$ $14$ $6$
One student of the class is chosen at random. What is the probability that the chosen students blood group is:
$i. O?$
$ii. AB?$
Answer
Total number of students $= 40$
$i.$ Number of students with blood group $O = 14$
Let $E_1$ be the event that the selected student's blood group is $O.$
$\therefore$ Required probability $= P(E_1) =\frac{14}{40}=0.35$
$ii.$ Number of students with blood group $AB = 6$
Let $E_2$ be the event that the selected student's blood group is $AB.$
$\therefore$ Required probability $= P(E_2) =\frac{6}{40}=0.15$
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Question 43 Marks
The table given below shows the marks obtained by $30$ students in a test.
Marks
$($Class interval$)$		 $1 - 10$ $11 - 20$ $21 - 30$ $31 - 40$ $41 - 50$
Number of students
$($Frequency$)$		 $7$ $10$ $6$ $4$ $3$
Out of these students, one is chosen at random. What is the probability that the marks of the chosen student:
$i.$ Are $30$ or less?
$ii.$ Are $31$ or more?
$iii.$ Lie in the interval $21 - 30$?
Answer
Total number of students $= 30$
$i.$ Probability that the marks of the chosen student are $30$ or less $=\frac{7+10+6}{30}=\frac{23}{30}$
$ii.$ Probability that the marks of the chosen student are $31$ or less $=\frac{4+3}{30}=\frac{7}{30}$
$iii.$ Probability that the marks of the chosen student lie in the interval $21 - 30 =\frac{6}{30}=\frac{1}{5}$
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Question 53 Marks
It is known that a box of $800$ electric bulbs contains $36$ defective bulbs.One bulb is taken at random out of the box. What is the probability that the bulb chosen is non-defective?
Answer
Total number of electric bulbs $= 800$
Number of defective bulbs $= 36$
$\Rightarrow $ Number of non-defective bulbs $= 800 - 36 = 764$
Hence, Probability that the bulb chosen is non-defective
$=\frac{\text{Number of non-defective bulbs}}{\text{Total number of bulbs}}$
$=\frac{764}{800}$
$=\frac{191}{200}$
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Question 63 Marks
In a survey of $200$ ladies, it was found that $142$ like coffee, while $58$ dislike it. Find the probability that a lady chosen at random:
$i.$ Likes coffee
$ii.$ Dislikes coffee.
Answer
Total number of tosses $= 200$ Number of ladies who likes coffee $= 142$ Number of ladies who dislikes coffee $= 58$
Let $E_1$ and $E_2$ be the events that the selected lady likes and dislikes coffee, respectively. Then,
$i. P($selected lady likes coffee$) = P(E_1) =\frac{\text{Number of ladies who like coffee}}{\text{Total number of laies}}$
$=\frac{142}{200}=0.71$
$ii. P($selected lady dislikes coffee$) = P(E_1) =\frac{\text{Number of ladies who dislike coffee}}{\text{Total number of laies}}$
$=\frac{58}{200}=0.29$
Remark: In the given survey, the only possible outcomes are $E_1$ and $E_2$ and $P(E_1) + P(E_2) = (0.71 + 0.29) = 1$.
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Question 73 Marks
A coin is tossed $500$ times and we get: heads: $285$ times and tails:$ 215$ times. When a coin is tossed at random, what is the probability of getting
$i.$ A head?
$ii.$ A tail?
Answer
Total number of tosses $= 500$ Number of heads $= 285$ Number of tails $= 215$
$i.$ Let $E$ be the event of getting a head.
$P($getting a head$) = \text{P(E)}=\frac{\text{Number of heads coming up}}{\text{Total number of trials}}$
$=\frac{285}{500}=0.57$
$ii.$ Let $F$ be the event of getting a tail.
$P($getting a tail$) = \text{P(E)}=\frac{\text{Number of heads coming up}}{\text{Total number of trials}}$
$=\frac{215}{500}=0.43$
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