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Maths 5 Marks Questions

Probability · Jharkhand Board (JAC) STD 7 (Jharkhand - English Medium). 3 questions.

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Question 15 Marks
A coin is tossed $300$ times and we get head $: 136$ times and tail $: 164$ times. When a coin is tossed at random, What is the probability of getting $(i)$ a head, $(ii)$ a tail$?$
Answer
Total number of trials $= 300$
Number of times a head is obtained $= 136$
Number of times a tail is obtained $= 164$
$i.$ Probability of getting head$=\frac{\text{Number of times heads is obtained}}{\text{Total number of trials}}=\frac{136}{300}=\frac{34}{75}$
$ii.$ Probability of getting tail$=\frac{\text{Number of times tails is obtained}}{\text{Total number of trials}}=\frac{164}{300}=\frac{41}{75}$
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Question 25 Marks
Two coins are tossed simultaneously $200$ times and we get two heads: $58$ times, one head: $83$ times: $0$ head: $59$ times. When two coins are tossed at random, what is the probability of getting $(i)\ 2$ heads, $(ii)\ 1$ head, $(iii)\ 0$ head$?$
Answer
Total number of trials $= 200$
Number of times $2$ heads are obtained $= 58$
Number of times one head is obtained $= 83$
Number of times no head is obtained $= 59$
$i.$ Probability of getting $2$ heads $=\frac{\text{Number of times 2 heads have been obtained}}{\text{Total number of trials}}=\frac{58}{200}=\frac{29}{100}$
$ii.$ Probability of getting $1$ head $=\frac{\text{Number of times 1 head has been obtained}}{\text{Total number of trials}}=\frac{83}{200}$
$iii.$ Probability of getting $0$ head $=\frac{\text{Number of times head has not been obtained}}{\text{Total number of trials}}=\frac{59}{200}$
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Question 35 Marks
A dice is thrown $100$ times and the outcomes are noted as given below:
Outcome
$1$
$2$
$3$
$4$
$5$
$6$
Frequency
$21$
$14$
$18$
$15$
$23$
$9$
When a dice is thrown at random, what is the probability of getting a $(i)\ 3,\ (ii)\ 6,\ (iii)\ 6,\ (iv) \ 1?$
Answer
Total number of trials $= 100\ N$
umber of times $3$ is obtained $= 18$
Number of times $6$ is obtained $= 9$
Number of times $4$ is obtained $= 15$
Number of times $1$ is obtained $= 21$
$i.$ Probability of getting a $ 3=\frac{\text{Number of times 3 is obtained}}{\text{Total number of trials}}=\frac{18}{100}=\frac{9}{50}$
$ii.$ Probability of getting a $6 =\frac{\text{Number of times 6 is obtained}}{\text{Total number of trials}}=\frac{9}{100}$
$iii.$ Probability of getting a $ 4 =\frac{\text{Number of times 4 is obtained}}{\text{Total number of trials}}=\frac{15}{100}=\frac{3}{20}$
$iv.$ Probability of getting a $1 =\frac{\text{Number of times 1 is obtained}}{\text{Total number of trials}}=\frac{21}{100}$
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