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Maths 4 Marks Questions

Probability · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 24 questions.

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Question 14 Marks
Tickets numbered 2, 3, 4, 5, ..., 100, 101 are placed in a box and mixed thoroughly. One ticket is drawn at random from the box. Find the probability that the number on the ticket is:
  1. An even number.
  2. A number less than 16.
  3. A number which is a perfect square.
  4. A prime number less than 40.
Answer
Total number of tickets = 100
  1. Even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 96, 98, 100
Tatal number of even number = 50
P(getting a even number)
$=\frac{50}{100}$
$=\frac{1}{2}$
  1. Numbers less than 16 are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15
Total number of number less than 16 is 14
P(getting a number less than 16)
$=\frac{14}{100}$
$=\frac{7}{50}$
  1. Numbers which are perfect square are 4, 9, 16, 25, 36, 49, 64, 81, 100
Total number of perfect squares = 9
P(getting a perfect square)
$=\frac{9}{100}$
  1. Prime number less than 40 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37
P(getting a prime number less 40)
$=\frac{12}{100}$
$=\frac{3}{25}$
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Question 24 Marks
A box contains cards bearing numbers $6$ to $70.$ If one card is drawn at random from the box, find the probability that it bears:
  1. A one-digit number.
  2. A number divisible by $5.$
  3. An odd number less than $30.$
  4. A composite number between $50$ and $70.$
Answer
To find the number of cards in the bag, we use the general formula for an $AP$ since the number on the cards are in $AP.$
Here, first term, $a = 6$, common difference $= d = 1$
Let n be the number of cards in the bag.
$a_n = a + (n - 1)d$
$\Rightarrow 70 = 6 + (n - 1)d$
$\Rightarrow 64 = n - 1$
$\Rightarrow n = 65$
So, there 65 cards in the bag.
  1. The one-digit number cards are $6, 7, 8$ and $9.$
So, there are $4$ possible outcomes.
$P($getting a one-digit numbered card$)$
$=\frac{4}{65}$
  1. The number divisible by $5$ are $10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65$ and $70.$
$P($Getting a number divisible by $5)$
$=\frac{13}{65}$
$=\frac{1}{5}$
  1. The odd number less than $30$ are $7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27$ and $29.$
$P($getting an odd number less than $30)$
$=\frac{12}{65}$
  1. There are $21$ number from $50$ to $70$.
The composite number between $50$ and $70$ are $51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69$
$P($getting a composite number between $50$ and $70)$
$=\frac{15}{65}$
$=\frac{3}{13}$
 
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Question 34 Marks
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing:
  1. An ace.
  2. A '4' cf spades.
  3. A '9' of a black suit.
  4. A red king.
Answer
Total number of all possible outcomes = 52
  1. P(getting an ace)
$=\frac{4}{52}$

$=\frac{1}{13}$
  1. P(getting a '4' of spades)
$=\frac{1}{52}$
  1. P(a '9' of black suit)
$=\frac{2}{52}$

$=\frac{1}{26}$
  1. P(getting a red king)
$=\frac{2}{52}$

$=\frac{1}{26}$
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Question 44 Marks
A box contains 25 cards numbered from 1 to 25. A card is drawn at random from the bag. Find the probability that the number on the drawn card is:
  1. Divisible by 2 or 3.
  2. A prime number.
Answer
There are 25 cards in total.
  1. The number divisible by 2 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.
The number divisible by 3 are 3, 6, 9, 12, 15, 18, 21 and 24.

So, the total number of possible outcomes = 16

Note that 6, 12, 18 and 24 are twice.

However, we count these number only once.

P(getting a number divisible by 2)

$=\frac{16}{25}$
  1. The primes are 2, 3, 5, 7, 11, 13, 17, 19 and 23.
So, there are 9 possible outcomes.

P(getting a prime)

$=\frac{9}{25}$
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Question 54 Marks
Two different dice are tossed together. Find the probability that
  1. The number on each die is even.
  2. The sum of numbers appearing on the two dice is 5.
Answer
If two dice are rolled together, the possible outcomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, there are 36 outcomes.
  1. The following outcomes have an even number on each dice.
(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)
So, there are 9 possible outcomes.
P(getting even numbers on each dice)
  1. The following outcomes give the sum 5.
(1, 4), (4, 1), (2, 3), (3, 2)
So, there are 4 possible outcomes.
P(getting a sum of 5)
$=\frac{4}{36}$
$=\frac{1}{9}$
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Question 64 Marks
Find the probability of getting the sum of two numbers, less than 3 or more than 11, when a pair of distinct dice is thrown together.
Answer
When two dice are thrown, the possible outcomes or Sample Space = {(1, 1), (1, 2), (1, 3), ..., (1, 6), (2, 1),(2, 2), ..., (2, 6), (3, 1), (3, 2), ..., (3, 6), (4, 1), ..., (4, 6), (5, 1) ..., (5, 6), (6, 1), ..., (6, 6)}
So there are total 36 outcomes which two dice are thrown.
According to question, we have to find the sum of numbers such that they are less than 3 and more than 11.
So, Let E denotes the number of outcomes for the required circumstance.
$\therefore\ \text{E}=\big\{{(1,1),(6,6)}\big\}$
Thus, number of outcomes, n(E) = 2
$\therefore\ $Required Probability
$=\frac{\text{n(E)}}{\text{S}}$
$=\frac{2}{36}$
$=\frac{1}{18}$
Hence, $\frac{1}{18}$ is the required probability.
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Question 74 Marks
Two dice are rolled once. Find the probability of getting such numbers on two dice whose product is a perfect square.
Answer
If two dice are rolled together, the possible oucomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, there are 36 outcomes.
The be a following outcomes will give the product to be a perfect square.
(1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)
So, there are 8 possible outcomes.
p(getting number whose product is a perfect square)
$=\frac{8}{36}$
$=\frac{2}{9}$
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Question 84 Marks
Two different dice are rolled simultanceously. Find the probability that the sum of the numbers on the two dice is 10.
Answer
If two dice are rolled together, the possible outcomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, there are 36 outcomes.
The following outcomes will give sum of numbers on the two di ce to be 10:
(2, 5), (5, 2) and (5, 5)
So, there are 3 possible outcomes.
P(getting numbers whose sum is 10)
$=\frac{3}{36}$
$=\frac{1}{12}$
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Question 94 Marks
The king, the jack and the 10 of spades are lost from a pack of 52 cards and a card is drawn from the remaining cards after shuffling. Find the probability of getting a:
  1. Red card.
  2. Black jack.
  3. Red king.
  4. 10 of hearts.
Answer
$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$
In a standard deck of 52 cards,
13 diamonds, 13 club, 13 hearts, 13 spade in which 12 face cards
The king, the jack and the 10 spades are lost from a pack of 52 cards
cards remaining are 52 - 1 - 1 - 10 = 40
  1. Red card
→ If the king and jack is from red card then number of red cards = 24

Fovorable outcome = 24

Total number of outcome = 40

Probability of a red card drawn is:

$\text{Probability}=\frac{24}{40}=\frac{3}{5}$

→ If the king and jack is not from red card then number of red cards = 26

Favorable oucome = 26

Total number of outcome = 40

Probability of red card drawn is:

$\text{Probability}=\frac{26}{40}=\frac{13}{20}$
  1. Black jack
→ If the jack is not from black card then number of black cards = 26

Favorable outcome = 26

Total number of outcome = 40

Probability of black jack drawn is:

$\text{Probability}=\frac{26}{40}=\frac{13}{20}$
  1. Red king
→ If the king is from red card then number of red cards = 25

Favorable outcome = 25

Total number of outcome = 40

Probability of red king drawn is:

$\text{Probability}=\frac{25}{40}=\frac{5}{8}$

→ If the king is not from red then number of red cards = 26

Favorable outcome = 26

Total number of outcome = 40

Probability of red king drawn is:

$\text{Probability}=\frac{26}{40}=\frac{13}{20}$
  1. 10 of hearts
→ If the king and jack is from hearts then number of hearts cards = 8

Favorable outcome = 8

Total number of outcome = 40

Probability of 10 of hearts drawn is:

$\text{Probability}=\frac{8}{40}=\frac{1}{5}$

→ If the king and jack is from hearts then number of hearts cards = 10

Fovorable outcome = 10

Total number of outcome = 40

Probability of 10 of hearts drawn is:

$\text{Probability}=\frac{10}{40}=\frac{1}{4}$
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Question 104 Marks
Two different dice are thrown together. Find the probability that the number obained have
  1. Even sum.
  2. Even prouduct.
Answer
(1, 1), (1, 2), (2, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
  1. Even sum outcome = (1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5,1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)
$\text{P}=\frac{18}{36}$
$=\frac{1}{2}$
  1. Even product = (1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4,1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5,2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6),
$\text{P}=\frac{27}{36}$
$=\frac{3}{4}$
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Question 114 Marks
One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:
  1. A king of red suit.
  2. A face card.
  3. A red face card.
  4. A queen of black suit.
  5. A jack of hearts.
  6. A spade.
Answer
We know that there are 52 cards in all.
Total number of oucomes = 52
  1. The number of red kings = 2
P(getting a king of red suit)

$=\frac{2}{52}$

$=\frac{1}{26}$
  1. The number of face cards = 12
P(getting a red face card)

$=\frac{12}{52}$

$=\frac{3}{13}$
  1. The number of red face cards = 6
P(getting a red face card)

$=\frac{6}{52}$

$=\frac{3}{26}$
  1. The number of queen of black suit = 2
P(getting a queen of black suit)

$=\frac{2}{52}$

$=\frac{1}{26}$
  1. The number of jack of hearts = 1
P(getting a jack of hearts)

$=\frac{1}{52}$
  1. The number of spade = 13
P(getting a spade)

$=\frac{13}{52}$

$=\frac{1}{4}$
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Question 124 Marks
Two different dice are thrown together. Find the probability that the numbers obtained:
  1. Have a sum less than 7.
  2. Have a product less than 16.
  3. Is a doublet of odd numbers.
Answer
The outcomes when two dice are thrown together are:
(1, 1), (1, 2), (2, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
  1. Let A be the event of getting the number whose sum is less than 7.
The outcomes in favour of event A are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2) and (5, 1)

Number of favourable outcones = 15

$\therefore\ \text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$=\frac{15}{36}$

$=\frac{5}{12}$
  1. Let B be the event of getting the number whose product is less than 16.
The outcomes in favour of event B are: (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1) and (6, 2).
  1. Let C be the event of getting the number which are doublets of odd numbers
The outcomes in favour of event C are (1, 1), (3, 3) and (5, 5)

Number of favourable outcomes = 3

$\therefore\ \text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$

$=\frac{3}{36}$

$=\frac{1}{12}$
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Question 134 Marks
Two different dice are thrown together. Find the probability that:
  1. The sum of the numbers appeared is less than 7.
  2. The product of the numbers appeared is less than 18.
Answer
If two dice are rolled together, the possible outcomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, there are 36 outcomes.
The their tops to be less than 7.
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), (5, 1).
So, there are 15 possible outcomes.
P(getting number whose sum of the tops to be less than 7)
$=\frac{15}{36}$
$=\frac{5}{12}$
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Question 144 Marks
Cards numbered 1 to 30 are put in a bag. A card is drawn at random from the bag. Find the probability that the number on the drawn card is:
  1. Not divisible by 3.
  2. A prime number greater that 7.
  3. Not a perfect square number.
Answer
There 30 cards in the bag.
  1. The number divisible by 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, and 30.
So, the number not divisible by 3 are 30 - 10 = 20

Thus, there are 20 posiible outcomes.

P(getting a number divisible by 3)

$=\frac{20}{30}$

$=\frac{2}{3}$
  1. The primes greater than 7 are 11, 13, 17, 19, 23 and 29.
So, there are 6 possible outcomes.

P(getting a prime greater than 3)

$=\frac{6}{30}$

$=\frac{1}{5}$
  1. The perfect squares would be 1, 4, 16 and 25.
So, there are 5 perfect squares, and hence 25

non-perfect squares.

P(getting a perfect square)

$=\frac{25}{30}$

$=\frac{5}{6}$
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Question 154 Marks
A die is rolled twice. Find the probability that:
5 will come up both the times.
Answer
5 will come up on both dice in 1 case = (5, 5)
$\therefore\ $Probability that 5 will come on both dice
$=\frac{1}{36}$
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Question 164 Marks
Two dice are rolled together. Find the probability of getting such numbers on two dice whose product is a perfect square.
Answer
If two dice are rolled together, the possible outcomes are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
So, there are 36 outcomes.
The following outcomes will give the product to be a perfect square.
(1, 1), (1, 4), (2, 2), (3, 3), (4, 1), (4, 4), (5, 5), (6, 6)
So, there are 8 possible outcomes.
P(getting number whose product is a perfect square)
$=\frac{8}{36}$
$=\frac{2}{9}$
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Question 174 Marks
A box contains cards numbered 3, 5, 7, 9, ..., 35, 37. A card is drawn at random from the box. Find the probability that the number on the card is a prime number.
Answer
The number on the cards are 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35 and 37.
So, there 18 cards.
The cards with primes are 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, and 37.
So, there are 11 possible outcomes.
P(getting a prime number)
$=\frac{11}{18}$
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Question 184 Marks
A card is drawn at random from a well-shuffled deck of playing deck of playing cards. Find the probability that the card drawn is:
  1. A card of spades or an ace.
  2. A red king.
  3. either a king or a queen.
  4. neither a king nor a queen.
Answer
Total number of cards = 52
  1. There are 13 cards of spade (including 1 ace) and 3 more ace cards are there in a pack of cards
$\therefore\ $P(getting a card of spades or an ace)

$=\frac{16}{52}$

$=\frac{4}{13}$
  1. There are 2 red kings in a pack of cards
$\therefore\ $P(getting a red king)

$=\frac{2}{52}$

$=\frac{1}{26}$
  1. There are 4 king and 4 queens in a pack of cards
$\therefore\ $P(getting either a king or a queen)

$=\frac{8}{52}$

$=\frac{2}{13}$
  1. P(getting neither a king nor a queen)
$=\Big(1-\frac{2}{13}\Big)$

$=\frac{11}{13}$
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Question 194 Marks
Cards bearing numbers $1, 3, 5, ..., 35$ are kept in a bag. A card is drawn at random from the bag. Find the probability of getting a card bearing:
  1. A prime number less than $15$.
  2. A number divisible by $3$ and $5$.
Answer
To find the number of cards in the bag, we use the general formula for an AP since the numbers on the cards are in AP.
Here, first term, $a = 1,$ common difference $= d = 2$
Let n be number of cards in the bag.
$a_n = a + (n - 1)d$
$\Rightarrow 35 = 1 + (n - 1)(2)$
$\Rightarrow 34 = 2n - 2$
$\Rightarrow n = 18$
So, there $18$ cards in the bag.
  1. The primes less than $15$ are $3, 5, 7, 11$ and $13$.
So, there are $6$ possible outcomes.
P(getting a prime less than $15$)
$=\frac{5}{18}$
  1. The number divisible by $3$ and $5$ is $15$ and $30$.
P(getting a $15$ or $30$)
$=\frac{2}{18}$
$=\frac{1}{9}$
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Question 204 Marks
A die is rolled twice. Find the probability that:
5 Will not come up either time.
Answer
Two dice are thrown simultaneouslyTotal number of outcomes = 66 = 36
Favourable cases are:
(1, 1), (1, 2), (1, 3), (1, 4), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 6)
= 25
$\therefore\ $Probability that 5 will not come upon either die
$=\frac{25}{36}$
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Question 214 Marks
A card is drawn at random from a well-shuffled deck of 52 cards. Find the probability of getting:
  1. A queen.
  2. A diamond.
  3. A king or an ace.
  4. A red ace.
Answer
Total number of cards = 52
  1. There are 4 queen cards in a pack of cards
$\therefore\ $Probability of getting a queen card

$=\frac{4}{52}$

$=\frac{1}{13}$
  1. There are 13 cards of diamond in a pack of cards
$\therefore\ $Probability of getting a diamond card

$=\frac{13}{52}$

$=\frac{1}{4}$
  1. In a pack of cards there are 4 kings and 4 aces Number of such cards = 4 + 4 = 8
Probability of getting eithera king or an ace

$=\frac{8}{52}$

$=\frac{2}{13}$
  1. There are two red aces in a pack of cards
$\therefore\ $Probability of getting a red ace

$=\frac{2}{52}$

$=\frac{1}{26}$
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Question 224 Marks
A die is rolled twice. Find the probability that:5 will come up exactly one time.
Answer
Favourable cases are: (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)
= 11
$\therefore\ $Probability that 5 will come at least once
$=\frac{11}{36}$
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Question 234 Marks
Peter throws two different dice together and finds the product of the two numbers obtained. Rina throws a die squares the number obtained. Who has the better chance to get the number 25?
Answer
Let us first write the all possible oucomes when Peter throws two different dice together.
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
$\therefore$ Total number of outcomes = 36
The favorable outcome for getting the product of numbers on the dice equal to 25 is (5, 5).
Favourable number of outcomes = 1
$\therefore$ Probability that Peter gets the product of numbers as 25
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}$
$=\frac{1}{36}$
The out comes when Rina throws a die are 1, 2, 3, 4, 5, 6.
$\therefore$ Total number of outcomes = 6
Rina throws a die and squares the number, so to get the number 25, the favourable outcomes is 5.
Favourable number of outcomes = 1
$\therefore$ Probability that Rina gets the square of numbers as 25
$=\frac{\text{Favourable number of outcomes}}{\text{Total number of outcomes}}$
$=\frac{1}{6}$
As, $\frac{1}{6}>\frac{1}{36},$ So Rina has better change to get the number 25.
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Question 244 Marks
Cards marked with numbers 5 to 50 are placed in a box and mixed thoroughly. A card us drawn from the box at random. Find the probability that the number on the taken out card is:
  1. A prime number less than 10.
  2. A number which is a perfect square.
Answer
There are 46 cards in total.
  1. The primes less than 10 are 5 and 7.
P(getting a prime less than 10)

$=\frac{2}{46}$

$=\frac{1}{23}$
  1. The perfect squares are 9, 16, 25, 36 and 49.
So, there are 5 possible outcomes.

p(getting a perfect square)

$=\frac{5}{46}$
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