5 Marks Questions

Question 15 Marks

Peter throws two different dice together and finds the product of the two numbers as 25. Rina throws a die and squares the number obtained. Who has the better chance to get the number

Answer

We know that the total number of outcome when two dice are thrown together is 36
Favorable out comes $=(5,5)$
Probability of getting two numbers having product as $25=\frac{1}{36}$
Rina throws the die only once. So, the total number of outcomes will be 6 only.
Favorable outcomes = (5)
Probability of getting a number whose square is $25=\frac{1}{6}$
Clearly, $\frac{1}{6}>\frac{1}{36}$
Hence, Rina has the better chance to get the number 25.

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Question 25 Marks

Two different dice are thrown together. Find the probability that the numbers obtained have
(i) Even sum, and
(ii) Even product.

Answer

The outcomes when two dices are thrown together
$\left\{\begin{array}{l}(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \$2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \$3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \$4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \$5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \$6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\end{array}\right\}$
There are 36 total outcomes
(i) When sum of numbers is even
Let $\text {B}$ be the event of getting even sum.
$\left\{\begin{array}{l}(1,1),(1,3),(1,5) \$2,2),(2,4),(2,6) \$3,1),(3,3),(3,5) \$4,2),(4,4),(4,6) \$5,1),(5,3),(5,5) \$6,2),(6,4),(6,6)\end{array}\right\}$
There are 18 favourable outcomes.
Probability for even sum outcomes
$P(A)=\frac{18}{36}=\frac{1}{2}$
(ii) Even product outcome
Let $\text {B}$ be the event of getting even product.

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Question 35 Marks

A bag contains cards numbered from 1 to 49. A card is drawn from the bag at random, after mixing the cards thoroughly. Find the probability that the number on the drawn card is:
(i) an odd number
(ii) a multiple of 5
(iii) a perfect square
(iv) an even prime number

Answer

(i) Total number of cards $=49$
Odd numbers from 1 to 49 are
$1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39,41,43,45,47,49$
Total number of favourable outcomes $=25$
Hence the require probability
$=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{25}{49}$
(ii) Sample space or total number of outcome = 49
Multiples of 5 that can be considered as the favourable number of outcomes are $5,10,15,20,25,30,35,40,45$
The number of favourable outcomes $=9$
Hence the required probability
$=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{9}{49}$
(iii) Sample space or total number of outcomes = 49
The numbers less than or equal to 49 that are perfect squares are $1,4,9,16,25,36,49$
Total number of favourable outcomes $=7$
Hence the required probability
$=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{7}{49}$
(iv) Sample space or total number of out 20 comes $=49$
Only one even prime number exists and that is, 2
Total number of favourable outcomes $=1$
Hence the required probability
$=\frac{\text { Total number of favourable outcomes }}{\text { Total possible outcomes }}=\frac{1}{49}$

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Question 45 Marks

Red queens and black jacks are removed from a pack of playing cards. A card is drawn at random from the remaining cards, after reshuffling them. Find the probability that the drawn card is
(i) a king
(ii) of red colour
(iii) a face card
(iv) a queen

Answer

Total cards in a pack $=52$
Number of black jacks $=2$
Number of red queens $=2$
Number of cards remaining after removing black jacks and red queens $=52-2-2=48$
(i) Probability that the card drawn is a king
$=\frac{\text { Number of kings }}{\text { Total number of cards }}$
Probability that the card drawn is a king
$=\frac{4}{48}=\frac{1}{12}$
The probability of drawing a king after removing
black jacks and red queens is $\frac{1}{12}$
(ii) Probability that the card drawn is of red color
$=\frac{\text { Number of red cards }}{\text { Total number of cards }} .$
There are 26 red cards in a pack out of which 2 red cards are removed
Number of red cards remaining $=26-2-24$
Probability that the card drawn is of red color
$=\frac{24}{48}=\frac{1}{2}$
The probability of drawing a red card after removing black jacks and red queens is $\frac{1}{2}$
(iii) Probability that the card drawn is a face card
$=\frac{\text { Number of face cards }}{\text { Total number of cards }}$
There are 12 face cards in a pack out of which 4 are removed.
Number of face cards remaining $=12-4=8$
Probability that the card drawn is a face card
$=\frac{8}{48}=\frac{1}{6}$
The probability of drawing a face card after removing black jacks and red queens is $\frac{1}{6}$
(iv) Probability that the card drawn is a queen
$=\frac{\text { Number of queens }}{\text { Total number of cards }}$
There are 4 queens in a pack of cards out of which 2 are removed.
Number of queens remaining $=4-2=2$
Probability that the card drawn is a queen
$=\frac{2}{48}=\frac{1}{24}$
The probability of drawing a queen after removing
black jacks and red queens is $\frac{1}{24}$.

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Question 55 Marks

A group consists of 12 persons, of which 3 are extremely patient, other 6 are extremely honest and rest are extremely kind. A person from the group is selected at random. Assuming that each person is equally likely to be selected, find the probability of selecting a person who is
(i) extremely patient
(ii) extremely kind or honest.
Which of the above values you prefer more?

Answer

Number of people in the group = 12
Number of possible outcomes $=12$
Let us assume, $E_1$ as an event for selecting extremely patient people and $E_2$ be the event for selecting extremely king or honest.
Number of outcomes for $E_1=3$
Number of extremely honest people $=6$
Extremely kind people $=12-(6+3)=3$
Number of outcomes for $E_2=9$
(i) P (Extremely Patient)
$=P\left(E_1\right)=\frac{\text { Outcomes for } E_1}{\text { Total possible outcomes }}=\frac{3}{12}=\frac{1}{4}$
(ii) P (Kind and Honest)
$=P\left(E_2\right)=\frac{\text { Outcomes for } E_2}{\text { Total possible outcomes }}=\frac{9}{12}=\frac{3}{4}$

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