2 Marks Questions

Question 12 Marks

When we toss a coin, there are two possible outcomes-Head or Tail. Therefore, the probability of each outcome is $\frac{1}{2}.$ Justify your answer.

Answer

There are two outcomes of equally in all manner. So probability of both head and tail are equal to $\frac{1}{2}$ each.
Hence, the given statement is true.

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Question 22 Marks

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Answer

Number of total outcomes = 36
When product of numbers appearing on them is less than 9, then possible ways are (1, 6), (1, 5) (1, 4), (1, 3), (1, 2), (1, 1), (2, 2), (2, 3), (2, 4), (3, 2), (4, 2), (4, 1), (3, 1), (5, 1), (6, 1) and (2, 1).
Number of possible ways = 16
$\therefore\ \text{Required probability}=\frac{16}{36}=\frac{4}{9}$

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Question 32 Marks

Box A contains 25 slips of which 19 are marked Re 1 and other are marked Rs. 5 each. Box B contains 50 slips of which 45 are marked Re 1 each and others are marked Rs. 13 each. Slips of both boxes are poured into a third box and resuffled. A slip is drawn at random. What is the probability that it is marked other than Re 1?

Answer

Total number of slips in a box, n(S) = 25 + 50 = 75
From the chart it is clear that, there are 11 slips which are marked other than Rs. 1.
$\therefore\ \ \text{Required probabitily}=\frac{\text{Number of slips other than Rs. 1}}{\text{Total number of slips}}=\frac{11}{75}$

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Question 42 Marks

Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

Answer

Not always. The median, mean and mode can be the same.
They may be equal if number of observations are odd and are equispaced.

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Question 52 Marks

Sushma tosses a coin $3$ times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

Answer

As the coin is' tossed $3$ times and gets tail each time but it is not necessary that $4^{th}$ time will be tail it may be either tail or head in any further toss.
Hence, the given statement is false.

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Question 62 Marks

A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) Fig. Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.

Answer

The area of region 3 is double either of 1 or 2 and area of 1 and 2 are equal so no. of outcomes (or probability) of region 3 is double of either 1 or 2.
So the outcomes of 1, 2, 3 are not equally likely to occur.

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Question 72 Marks

If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the $4^{\text {th }}$ toss? Give reason in support of your answer.

Answer

As we know that a coin has two equal chances always either head or tail. So next time on tossing he can get either tail or head.
So, the given statement is false.

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Question 82 Marks

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula $\bar{\text{x}}=\text{a}+\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}$ where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

Answer

Not always. Assumed mean can be considered any convenient number which makes calculation easy.

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Question 92 Marks

There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs. 100 each, 100 of them contain a cash prize of Rs. 50 each and 200 of them contain a cash prize of Rs. 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize?

Answer

Total number of sealed envelopes in a box, n(S) = 1000
Number of envelopes containing cash prize = 10 + 100 + 200 = 310
Number of envelopes containing no cash prize,
n(E) = 1000 - 310 = 690
$\therefore\ \ \text{P(E)}=\frac{\text{n(E)}}{\text{n(S)}}=\frac{690}{1000}=\frac{69}{100}=0.69$

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Question 102 Marks

A coin is tossed $3$ times. List the possible outcomes. Find the probability of getting:

All heads.
At least 2 heads.

Answer

The possible outcomes if a coin is tossed 3 times is, $S =\{( HHH ),( TTT ),( HTT ),( THT ),( TEH ),( THH ),( HTH ),( HHT )\}$
i. Let $E _1=$ Event of getting all head $=\{( HHH )\}$
$\therefore n\left(E_1\right)=1$
$\therefore P\left(E_1\right)=\frac{n\left(E_1\right)}{n(S)}=\frac{1}{8}$
ii. Let $E _2=$ Event of getting atieast 2 heads $=\{( HHT ),( HTH ),( THH ),( HHH )\}$
$\therefore n\left(E_2\right)=4$
$\therefore P\left(E_2\right)=\frac{n\left(E_2\right)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

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Question 112 Marks

A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and the probability of getting ‘not 1’ each is equal to $\frac{1}{2}.$ Is this correct? Give reasons.

Answer

A dice can be thrown in 6 different equally likely ways. Possible outcomes are given by S = {1, 2, 3, 4, 5, 6}.
P(getting 1) $=\frac{1}{6}$ and P(not getting 1) $=\frac{5}{6}.$
Hence, the given statement is not correct.

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Question 122 Marks

In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is $\frac{1}{4}.$ Is this correct? Justify your answer

Answer

False: In a family of three children events are (b, b, b,), (g, b, b), (g, g, b), (g, g, g) T(E) = 4
The probability of each is not $\frac{1}{4},$ because the outcomes are not equally likely.

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