4 Marks Questions

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:

An even number.
A number less than $16.$
A number which is a perfect square.
A prime number less than $40.$

Answer

Total number of tickets $= 100$

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)
$=\frac{50}{100}$
$=\frac{1}{2}$

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 $16)$
$=\frac{14}{100}$
$=\frac{7}{50}$

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)
$=\frac{9}{100}$

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:

A one$-$digit number.
A number divisible by $5.$
An odd number less than $30.$
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.

The one$-$digit number cards are $6, 7, 8$ and $9.$

So, there are $4$ possible outcomes.
P$($getting a one-digit numbered card$)$
$=\frac{4}{65}$

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 $5)$
$=\frac{13}{65}$
$=\frac{1}{5}$

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 $30)$
$=\frac{12}{65}$

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:

An ace.
A $'4'$ cf spades.
A $'9'$ of a black suit.
A red king.

Answer

Total number of all possible outcomes $= 52$

P$($getting an ace$)$

$=\frac{4}{52}$
$=\frac{1}{13}$

P$($getting a $'4'$ of spades$)$

$=\frac{1}{52}$

P$($a $'9'$ of black suit$)$

$=\frac{2}{52}$
$=\frac{1}{26}$

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:

Divisible by $2$ or $3.$
A prime number.

Answer

There are 25 cards in total.

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 $2)$
$=\frac{16}{25}$

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

The number on each die is even.
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.

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)

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:

Red card.
Black jack.
Red king.
$10$ 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$
i. Red card
$\rightarrow$ 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:
Probability $=\frac{24}{40}=\frac{3}{5}$
$\rightarrow$ 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:
Probability $=\frac{26}{40}=\frac{13}{20}$
ii. Black jack
$\rightarrow$ 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:
Probability $=\frac{26}{40}=\frac{13}{20}$
iii. Red king
$\rightarrow$ 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:
Probability $=\frac{25}{40}=\frac{5}{8}$
$\rightarrow$ 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:
Probability $=\frac{26}{40}=\frac{13}{20}$
iv. $10$ of hearts
$\rightarrow$ 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:
Probability $=\frac{8}{40}=\frac{1}{5}$
$\rightarrow$ 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:
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

Even sum.
Even prouduct.

Answer

$(1, 1), (1, 2), (2, 3), (1, 4), (1, 5), (1, 6)$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

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)$

$\text{P}=\frac{18}{36}$
$=\frac{1}{2}$

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:

A king of red suit.
A face card.
A red face card.
A queen of black suit.
A jack of hearts.
A spade.

Answer

We know that there are $52$ cards in all.
Total number of oucomes $= 52$

The number of red kings $= 2$

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

The number of face cards $= 12$

P(getting a red face card)
$=\frac{12}{52}$
$=\frac{3}{13}$

The number of red face cards $= 6$

P(getting a red face card)
$=\frac{6}{52}$
$=\frac{3}{26}$

The number of queen of black suit $= 2$

P(getting a queen of black suit)
$=\frac{2}{52}$
$=\frac{1}{26}$

The number of jack of hearts $= 1$

P(getting a jack of hearts)
$=\frac{1}{52}$

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:

Have a sum less than $7.$
Have a product less than $16.$
Is a doublet of odd numbers.

Answer

The outcomes when two dice are thrown together are:

$(1, 1), (1, 2), (2, 3), (1, 4), (1, 5), (1, 6)$

$(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)$

$(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)$

$(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)$

$(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)$

$(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$

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 $= 15$
$\therefore\ \text{P(A)}=\frac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$
$=\frac{15}{36}$
$=\frac{5}{12}$

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 $(6, 2).$

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:

The sum of the numbers appeared is less than $7.$
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:

Not divisible by $3.$
A prime number greater that $7.$
Not a perfect square number.

Answer

There 30 cards in the bag.

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 $3)$
$=\frac{20}{30}$
$=\frac{2}{3}$

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 $3)$
$=\frac{6}{30}$
$=\frac{1}{5}$

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:

A card of spades or an ace.
A red king.
either a king or a queen.
neither a king nor a queen.

Answer

Total number of cards $= 52$

There are $13$ cards of spade $($including $1$ ace$)$ and $3$ more ace cards are there in a pack of cards

$\therefore\ $P(getting a card of spades or an ace$)$
$=\frac{16}{52}$
$=\frac{4}{13}$

There are $2$ red kings in a pack of cards

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

There are $4$ king and $4$ queens in a pack of cards

$\therefore\ $P$($getting either a king or a queen$)$
$=\frac{8}{52}$
$=\frac{2}{13}$

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:

A prime number less than $15.$
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.
$a_n= a + (n - 1)d$
$⇒ 35 = 1 + (n - 1)(2)$
$⇒ 34 = 2n - 2$
$⇒ n = 18$
So, there $18$ cards in the bag.

The primes less than $15$ are $3, 5, 7, 11$ and $13.$

So, there are $6$ possible outcomes.
P(getting a prime less than $15)$
$=\frac{5}{18}$

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 simultaneously
Total 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:

A queen.
A diamond.
A king or an ace.
A red ace.

Answer

Total number of cards $= 52$

There are $4$ queen cards in a pack of cards

$\therefore\ $Probability of getting a queen card
$=\frac{4}{52}$
$=\frac{1}{13}$

There are $13$ cards of diamond in a pack of cards

$\therefore\ $Probability of getting a diamond card
$=\frac{13}{52}$
$=\frac{1}{4}$

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
$=\frac{8}{52}$
$=\frac{2}{13}$

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:

A prime number less than $10.$
A number which is a perfect square.

Answer

There are $46$ cards in total.

The primes less than $10$ are $5$ and $7.$

$P($getting a prime less than $10)$
$=\frac{2}{46}$
$=\frac{1}{23}$

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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