Question

Q1) peter throws 2 different dies together and finds the product if the 2 numbers obtained. Reena throws a die and squares the number obtained. Who has the better chance to get the no 25?

Q2) cars on which the numbers 1-100 are written are put in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the card taken out has (i) an even no. (ii) a number which is a multiple of 13 (iii) a perfect square number (iv) a prime no less than 20

Q3) the king, queen and jack of clubs are removed from a deck if 52 cards. The remaining cards are mixed together and then a card is drawn at random from it. Find the probability of getting (i) a face card (ii) a card of hearts (iii) a card of clubs (iv) a queen of diamonds

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Answer 1

Step-by-step explanation:

Q1) peter throws 2 different dies together and finds the product if the 2 numbers obtained. Reena throws a die and squares the number obtained. Who has the better chance to get the no 25?

Q2) cars on which the numbers 1-100 are written are put in a bag and mixed thoroughly. A card is drawn at random from the bag. Find the probability that the card taken out has (i) an even no. (ii) a number which is a multiple of 13 (iii) a perfect square number (iv) a prime no less than 20

Q3) the king, queen and jack of clubs are removed from a deck if 52 cards. The remaining cards are mixed together and then a card is drawn at random from it. Find the probability of getting (i) a face card (ii) a card of hearts (iii) a card of clubs (iv) a

it's very easy

Answer 2

Solution 1:

It is given that Peter throws 2 dice at a time and he multiplies both the numbers which are obtained. The possibile outcomes when 2 dice are thrown at a time 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)

These all are 36 outcomes

If Peter is Multiplying both the obtained number, then the possibility of getting 25 will be only in 1 case [which is (5,5)].

So,

$\leadsto\textsf{Peter's possibility of getting 25 = \dfrac{\sf Possible \;outcome}{\sf Total\; outcomes} }$

$\leadsto\boxed{\textsf{Peter's possibility of getting 25 = \dfrac{\bf 1}{\bf 36} }}$

Now, it is given that  Reena throws only 1 dice and squares the number which is obtained (Number²). Total possible outcomes when 1 dice is thrown are:-

1,2,3,4,5 and 6.

Possibility of getting 25 after squaring any of the number is possible only in 1 case(5²=25).

$\leadsto\textsf{Reena's possibility of getting 25 = \dfrac{\sf Possible \;outcome}{\sf Total\; outcomes} }$

$\leadsto\boxed{\textsf{Reena's possibility of getting 25 = \dfrac{\bf 1}{\bf 6} }}$

$\underline{\boxed{\sf\dfrac{1}{36}<\dfrac{1}{6}}}$

Hence possibility of Reena is more than that of Peter's.

Solution 2:

In this question, we are given that numbers from 1 to 100 are written on a card and put into a bag. We are asked to find the possibility of obtaining:

• An even number.

• Multiply of 13.

• Perfect square number.

• Prime number less than 20.

There are total outcomes=100 where 50 are even and 50 are odd. Even and odd number will be obtained after every proceeding number.

So,

$\leadsto\textsf{Possibility of obtaining even number = \dfrac{\sf Possible\: outcome}{\sf Total\: outcomes} }$

$\leadsto\textsf{Possibility of obtaining even number = \dfrac{\sf 50}{\sf 100} }$

$\leadsto\boxed{\textsf{Possibility of obtaining even number = \dfrac{\sf 1}{\sf 2} }}$

Multiples of 13 upto 100 are:

13,26,39,52,65,78,91

These are 7 outcomes.

$\leadsto\textsf{Possibility of obtaining multiply of 13= \dfrac{\sf Possible\: outcome}{\sf Total\: outcomes} }$

$\leadsto\boxed{\textsf{Possibility of obtaining multiply of 13= \dfrac{\sf 7}{\sf 100} }}$

Perfect square number upto 100 are:

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

These are 9 outcomes.

So,

$\leadsto\boxed{\textsf{Possibility of obtaining perfect square number= \dfrac{\sf 9}{\sf 100} }}$

Prime numbers less than 20 are:

2,3,5,7,9,11,13,17,19

These are 9 outcomes.

So,

$\leadsto\boxed{\textsf{Possibility of obtaining prime number less than 20= \dfrac{\sf 9}{\sf 100} }}$

Solution 3:

Cards in a deck are:

$\boxed{\begin{array}{c |c |c| c}\bf Heart&\bf Spade& \bf Club&\bf Diamond&&&&&\sf A&\sf A&\sf A&\sf A\\\sf 2&\sf 2&\sf 2&\sf 2\\\sf 3&\sf 3&\sf 3&\sf 3&\sf 4&\sf 4&\sf 4&\sf 4&\sf 5&\sf 5&\sf 5&\sf 5&\sf 6&\sf 6&\sf 6&\sf 6&\sf 7&\sf 7&\sf 7&\sf 7&\sf 8&\sf 8&\sf 8&\sf 8&\sf 9&\sf 9&\sf 9&\sf 9&\sf 10&\sf 10&\sf 10&\sf 10&\sf J&\sf J&\sf J&\sf J&\sf Q&\sf Q&\sf Q&\sf Q&\sf K&\sf K&\sf K&\sf K\end{array}}$

These are total 52 cards, of which face cards are of 3 types  J, Q, K. There are 13 cards of each symbol (Heart, Spade, Club, diamond).

In the Question we are given a statement that king(K), Queen (Q) and Jack(J) of club are removed from the deck. So total cards will remain 49.

Now we are asked to find probability of:

• A face card

• A card of heart

• A card of club

• A queen of diamond

~A face card

Total face card will be 3×4 (Symbols)=12

3 card are removed so face card will be 12-3=9

So,

$\leadsto\textsf{Possibility of obtaining a face card= \dfrac{\sf Possible\: outcome}{\sf Total\: outcomes} }$

$\leadsto\boxed{\textsf{Possibility of obtaining a face card= \dfrac{\sf 9}{\sf 49} }}$

~A card of heart

Each symbol has 13 cards.

So,

$\leadsto\boxed{\textsf{Possibility of obtaining a card of heart= \dfrac{\sf 13}{\sf 49} }}$

~A card of club

There are 13 cards of each symbol but 3 cards from club are removed. So total card of club will be 13-3=10

So,

$\leadsto\boxed{\textsf{Possibility of obtaining a card of club= \dfrac{\sf 10}{\sf 49} }}$

~A Queen of diamond

There is only 1 card of each symbol as given in the above table.

So,

$\leadsto\boxed{\textsf{Possibility of obtaining a Queen of diamond= \dfrac{\sf 1}{\sf 49} }}$

$\underline{\boxed{\pmb{\sf{\red{ Hope\:it\: helps\: :)}}}}}$