5 Marks Questions

Question 15 Marks

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

a king of red colour
a face card
a red face card
the jack of hearts
a spade
the queen of diamonds

Answer

Total number of cards in one deck of cards is $52 . \therefore$ Total number of outcomes $n =52$
i. Let $E_1=$ Event of getting a king of red color. So number of outcomes favourable to $E_1 m=2$ So $P\left(E_1\right)=$ $\frac{m}{n}=\frac{2}{52}=\frac{1}{26}$
ii. Let $E _2=$ Event of getting a face card
$\therefore$ Numbers of outcomes favourable to $E _2, m=12$. Hence $P \left( E _2\right)=\frac{m}{n}=\frac{12}{52}=\frac{3}{13}$
iii. Let $E_3=$ Event of getting a red face card
$\therefore$ Numbers of outcomes favourable to $E_3=6[\because$ there are 6 red face cards in a deck $]$ Hence $P\left(E_3\right)=$ $\frac{m}{n}=\frac{6}{52}=\frac{3}{26}$
iv. Let $E _4=$ Event of getting a jack of heart
$\therefore$ Numbers of outcomes favourable to $E_4=1[\because$ there is only one jack of heart in deck of cards.] Hence $P\left(E_4\right)=\frac{m}{n}=\frac{1}{52}$
v. Let $E _5=$ Event of getting a spade
$\therefore$ Numbers of outcomes favourable to $E _5=13[\because$ there are 13 spade in a deck $]$
Hence $P\left(E_5\right)=\frac{m}{n}=\frac{13}{52}$
vi. Let $E_6=$ Event of getting the queen of diamond
$\therefore$ Numbers of outcomes favourable to $E_6=1[\because$ there is only one queen of diamond in a deck] Hence, $P \left( E _6\right)=\frac{m}{n}=\frac{1}{52}$

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Question 25 Marks

Two dice, one blue and one grey, are thrown at the same time.Complete the following table:
Event: Sum on 2 dice 2 3 4 5 6 7 8 9 10 11 12
Probability $\frac{1}{{36}}$ $\frac{5}{{36}}$ $\frac{1}{{36}}$
A student argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability $\frac{1}{11}$. Do you agree with this argument? Justify your answer.

Answer

Total no. of possible outcomes when 2 dice are thrown $=6 \times 6=36$
Probability, $P(E)=\frac{\text { No. of favourable outcomes }}{\text { Total no. of possible outcomes }}$

It can be observed that,
To get the sum as 2, possible outcomes = (1, 1)
$P ( E )=\frac{1}{36}$
To get the sum as 3, possible outcomes = (2, 1) and (1, 2)
$P ( E )=\frac{2}{36}$
To get the sum as 4, possible outcomes = (3, 1), (1, 3), (2, 2)
$P ( E )=\frac{3}{36}$
To get the sum as 5, possible outcomes = (4, 1), (1, 4), (2, 3), (3, 2)
$P ( E )=\frac{4}{36}$
To get the sum as 6, possible outcomes = (5, 1), (1, 5), (2, 4), (4, 2), (3, 3)
$P ( E )=\frac{5}{36}$
To get the sum as 7, possible outcomes = (6, 1), (1, 6), (2, 5), (5, 2), (3, 4), (4, 3)
$P ( E )=\frac{6}{36}$
To get the sum as 8, possible outcomes = (6, 2), (2, 6), (3, 5), (5, 3), (4, 4)
$P ( E )=\frac{5}{36}$
To get the sum as 9, possible outcomes = (3, 6), (6, 3), (4, 5), (5, 4)
$P ( E )=\frac{4}{36}$
To get the sum as 10, possible outcomes = (4, 6), (6, 4), (5, 5)
$P ( E )=\frac{3}{36}$
To get the sum as 11, possible outcomes = (5, 6), (6, 5)
$P ( E )=\frac{2}{36}$
To get the sum as 12, possible outcomes = (6, 6)
$P ( E )=\frac{1}{36}$

Event: Sum on 2 dice	2	3	4	5	6	7	8	9	10	11	12
Probability	$\frac{1}{{36}}$	$\frac{2}{{36}}$	$\frac{3}{{36}}$	$\frac{4}{{36}}$	$\frac{5}{{36}}$	$\frac{6}{{36}}$	$\frac{5}{{36}}$	$\frac{4}{{36}}$	$\frac{3}{{36}}$	$\frac{2}{{36}}$	$\frac{1}{{36}}$

The probability of each of these sums will not be$\frac{1}{11}$as their sums are not equally likely.

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