[1 Mark Question Answer]

Question 11 Mark

Namita tossed a coin once. What is the probability of getting tail?

Answer

Sample space S = {H, T}
n(S) = 2
Event = {tail}
n(E) = 1
P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{1}{2}$

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Question 21 Mark

Namita tossed a coin once. What is the probability of getting Head?

Answer

Sample space
S = {H, T}
n(S) = 2.
Event = {Head}
n(E) = 1
$P(E)=\frac{n(E)}{n(S)}=\frac{1}{2}$.

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Question 31 Mark

A coin is tossed 100 times with the following frequency:
Head = 55, Tail = 45
find the probability of getting tail.

Answer

Event = {tail}
n(E) = 45
∴ Probability of event = $\frac{ n ( E )}{ n ( S )}=\frac{45}{100}=\frac{9}{20}$.

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Question 41 Mark

A coin is tossed 100 times with the following frequency:
Head = 55, Tail = 45.
Find the probability of getting head.

Answer

Event = {Head}
n(E) = 55
$\therefore$ Probability of event $=\frac{ n ( E )}{ n ( S )}=\frac{55}{100}=\frac{11}{20}$.

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Question 51 Mark

1000 tickets of a lottery were sold and there are 5 prizes on these tickets. If Namita has purchased one lottery ticket, what is the probability of winning a prize?

Answer

n(S) = 1000
n(E) = 5
P(E) = ?
$P(E)=\frac{n(E)}{n(S)}=\frac{5}{1000}=0.005$

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Question 61 Mark

It is known that a bax of 600 electric bulbs contain 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?

Answer

Number of non-defective bulbs = 600 - 12 = 588.
n(E) = 588
n(S) = 600
P(E) = ?
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{588}{600}=0.98$.

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Question 71 Mark

In a cricket match a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit the boundary?

Answer

n(S) = 30
E = { not hitting the boundary }
n(E) = 30 - 6 = 24
Probability of not hitting a boundary
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{24}{30}=\frac{4}{5}$

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Question 81 Mark

If the probability of winning a 5 game is 5/11. What is the probability of losing?

Answer

$P(E)=\frac{5}{11}$
$\therefore$ Probability of lossing $P(\bar{E})=1-P(E)=1-\frac{5}{11}=\frac{6}{11}$

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Question 91 Mark

Two dice are thrown simultaneously. Find the probability of getting six as the product.

Answer

n(S) = 36
Event = {getting 6 as a product}
= {(1,6), (2,3), (3,2), (6,1)}
n(E) = 4
∴ P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{4}{36}=\frac{1}{9}$

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Question 101 Mark

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card is drawn is ’10’ of a black suit.

Answer

Event = { '10' of blacksuit}
n(E) = 2
P(E) = ?
P(E) = $\frac{ n ( E )}{ n ( s )}=\frac{2}{52}=\frac{1}{26}$.

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Question 111 Mark

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is ‘2’ of spade

Answer

Event = { '2' of spade }
n(E) = 1
P(E) = ?
P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{1}{52}$

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Question 121 Mark

One card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a red face card

Answer

Total no. of possible outcomes = 52 (52 cards)
E ⟶ event of getting red face card
No. favourable outcomes = 6 { kings, queens, jacks of hearts & diamonds}
$P(E)_t=\frac{\text { No.of favorable outcomes }}{\text { Total no.of possible outcomes }}$
$P(E)=\frac{6}{52}=\frac{3}{26}$

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Question 131 Mark

A card is drawn from a pack of well-shuffled 52 playing cards. Find the probability that the card is drawn is a face card.

Answer

There are 52 playing cards in a pack.
Number of face cards = 12
P(drawing a face card) = $\frac{\text { Favourable outcomes }}{\text { Total outcomes }}=\frac{12}{52}=\frac{3}{13}$

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Question 141 Mark

One card is randomly drawn from a pack of 52 cards. Find the probability that: the drawn card is red and a king.

Answer

In randomly drawing a card from 52 cards.
n(s) = 52
Let C denote the event that the drawn card is red and a king.
n(C) = 2
P(C) = ?
n(S) = 52
∴ P(C) = $\frac{ n ( C )}{ n ( S )}=\frac{2}{52}=\frac{1}{26}$.

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Question 151 Mark

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card is drawn is either red or king.

Answer

n(S) = 52
Event = {either red or king}
Event = { 26 red cards + 2 kings }
Event = 28
n(E) = 28
P(E) = ?
∴ P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{28}{52}=\frac{7}{13}$

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Question 161 Mark

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card is drawn is red.

Answer

n(S) = 52
Event = {Red Cards}
n(E) = 26
P(E) = ?
$\therefore P(E)=\frac{n(E)}{n(S)}=\frac{26}{52}=\frac{1}{2}$.

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Question 171 Mark

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card is drawn is An ace.

Answer

n(S) = 52
Event = [ an ace ]
n(E) = 4
P(E) = ?
∴ P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{4}{52}=\frac{1}{13}$

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Question 181 Mark

One card is randomly drawn from a pack of 52 cards. Find the probability that: the drawn card is red or king.

Answer

In randomly drawing a card from 52 cards.
n(S) = 52
Let D denote the event that the drawn card is red or a king.
n(D) = 26 (red cards ) + 2(kings)
n(D) = 28
P(D) = $\frac{ n ( D )}{ n ( S )}=\frac{28}{52}=\frac{7}{13}$

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Question 191 Mark

One card is randomly drawn from a pack of 52 cards. Find the probability that: the drawn card is red and a king.

Answer

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Question 201 Mark

One card is randomly drawn from a pack of 52 cards. Find the probability that: the drawn card is an ace.

Answer

In randomly drawing a card from 52 cards.
n(S) = 52.
Let B denote the event that drawn card is an ace.
n(B) = 4
n(S) = 52
P(B) = ?
∴ P(B) = $\frac{ n ( B )}{ n ( S )}=\frac{4}{52}=\frac{1}{13}$

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Question 211 Mark

One card is randomly drawn from a pack of 52 cards. Find the probability that: the drawn card is red.

Answer

In randomly drawing a card from 52 cards.
n(S) = 52
Let A denote the event that the drawn card is red.
n(A) = 26
∴ P(A) = $\frac{ n ( A )}{ n ( S )}=\frac{26}{52}=\frac{1}{2}$

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Question 221 Mark

A dice is thrown once. What is the probability that the number is greater than 2?

Answer

Event = {number is greater than 2}
Event = {3, 4, 5, 6}
n(E) = 4
P(E) = ?
P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{4}{6}=\frac{2}{3}$.

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Question 231 Mark

A dice is thrown once. What is the probability that the number is even?

Answer

n(S) = 6
Event = {Even number}
Event = {2, 4, 6}
n(E) = 3
P(E) = ?
P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{3}{6}=\frac{1}{2}$

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Question 241 Mark

1800 families with 2 children were selected randomly and the following data were recorded:

No. of girls in a family	2	1	0
No. of families	700	850	250

Computer the probability of a family chosen at random having: No girls.

Answer

Total number of families n(S) = 1800.
Event = {No girl}
n(E) = 250
P(E) = ?
P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{250}{1800}=\frac{5}{36}$

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Question 251 Mark

1800 families with 2 children were selected randomly and the following data were recorded:

No. of girls in a family	2	1	0
No. of families	700	850	250

Computer the probability of a family chosen at random having: 2 girls

Answer

Total number of families n(S) = 1800.
Event = {2 girls}
n(E) = 700
P(E) = ?
∴ P(E) = $\frac{ n ( E )}{ n ( S )}=\frac{700}{1800}=\frac{7}{18}$

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Question 261 Mark

A die has 6 faces marked by the given numbers as shown below:

- 1

- 2

- 3

The die is thrown once. What is the probability of getting the smallest integer?

Answer

-1

-2

-3

Total number of outcomes = 6.
P( the smallest integer ) = $\frac{1}{6}$.

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Question 271 Mark

A die has 6 faces marked by the given numbers as shown below:

- 1

- 2

- 3

The die is thrown once. What is the probability of getting an integer greater than – 3.

Answer

-1

-2

-3

Total number of outcomes = 6.
P( an integer greater than - 3) =$\frac{5}{6}$.

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Question 281 Mark

A die has 6 faces marked by the given numbers as shown below:

- 1

- 2

- 3

The die is thrown once. What is the probability of getting a positive integer?

Answer

-1

-2

-3

Total number of outcomes = 6.
P ( a positive integer ) = $\frac{3}{6}=\frac{1}{2}$.

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Question 291 Mark

Two coins are tossed once. Find the probability of getting at least 1 tail.

Answer

If two coins are tossed once, then
S = {HH, HT, TH, TT}
⇒ N(S) = 4.
At least one tail
∴ Favourable outcome = 3
Required probability $=\frac{3}{4}$.

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Question 301 Mark

Two coins are tossed once. Find the probability of getting 2 heads.

Answer

If two coins are tossed once, then
S = {HH, HT, TH, TT}
⇒ N(S) = 4.
E: getting two heads
∴ N(E) = 1
$\therefore P(E)=\frac{N(E)}{N(S)}=\frac{1}{4}$

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Question 311 Mark

An unbiased dice is thrown. What is the probability of getting a number other than 4.

Answer

Sample Space = {1, 2, 3, 4, 5, 6}
n(S) = 6
Event = { Other than 4 }
= {1, 2, 3, 5, 6}
n(E) = 5
$\therefore P(E) \frac{n(E)}{n(S)}=\frac{5}{6}$.

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