# One Card is drawn at random from a well-shuffled deck of 52 cards. In how many of the following cases are the events E and F independent?

i. $E$: ‘the card drawn is spade’

$F$: ‘the card drawn is an ace’

ii. $E$: ‘the card drawn is black’

$F$: ‘the card drawn is a king ’

iii. $E$: ‘the card drawn is a king or queen’

$F$: ‘the card drawn is a queen or jack’

Answer

Verified

363k+ views

Hint: Here, in this solution we will use the concept of independent events i.e.., “the two events E and F are said to be independent if and only if $P(E \cap F) = P(E).P(F)$.”

Complete step-by-step answer:

i. Given,

E: ‘the card drawn is spade’

F: ‘the card drawn is an ace’

In a deck of 52 cards, 13 cards will be spade and 4 cards are ace and only 1 card is an ace of spades.

Therefore,

$P(E)$=$\dfrac{{13}}{{52}} = \dfrac{1}{4}$i.e.., the probability of drawing a spade from a deck of 52 cards.

$P(F) = \dfrac{4}{{52}} = \dfrac{1}{{13}}$i.e.., the probability of drawing an ace from a deck of 52 cards.

$P(E \cap F) = \dfrac{1}{{52}}$ i.e.., the probability of drawing a card which is spade as well as an ace from a deck of 52 cards.

Now, to prove $E$ and $F$events to be independent the following condition to be satisfied.

$P(E \cap F) = P(E).P(F) \to (1)$

So let us substitute the obtained values of$P(E)$, $P(F)$and $P(E \cap F)$in equation (1) we get,

$

\Rightarrow \dfrac{1}{{52}} = \dfrac{1}{4}.\dfrac{1}{{13}} \\

\Rightarrow \dfrac{1}{{52}} = \dfrac{1}{{52}}[\therefore L.H.S = R.H.S] \\

$

Therefore, we can say the events $E$ and $F$ are independent as they are satisfying the condition of independent events.

ii. Given,

$E$: ‘the card drawn is black’

$F$: ‘the card drawn is a king ’

In a deck of 52 cards, 26 cards are black and 4 cards are kings only 2cards are black as well as kings.

Therefore,

$P(E)$=$\dfrac{{26}}{{52}} = \dfrac{1}{2}$i.e.., the probability of drawing a black card from a deck of 52 cards.

$P(F) = \dfrac{4}{{52}} = \dfrac{1}{{13}}$i.e.., the probability of drawing a king from a deck of 52 cards.

$P(E \cap F) = \dfrac{2}{{52}} = \dfrac{1}{{26}}$ i.e., the probability of drawing a card which is black as well as king from a deck of 52 cards.

Now, to prove $E$ and $F$events to be independent the following condition to be satisfied.

$P(E \cap F) = P(E).P(F) \to (1)$

So let us substitute the obtained values of$P(E)$, $P(F)$and $P(E \cap F)$in equation (1) we get,

$

\Rightarrow \dfrac{1}{{26}} = \dfrac{1}{2}.\dfrac{1}{{13}} \\

\Rightarrow \dfrac{1}{{26}} = \dfrac{1}{{26}}[\therefore L.H.S = R.H.S] \\

$

Therefore, we can say the events $E$and $F$ are independent as they are satisfying the condition of independent events.

iii. Given,

$E$: ‘the card drawn is a king or queen’

$F$: ‘the card drawn is a queen or jack’

In a deck of 52 cards, 4 cards are kings, 4 cards are queens and 4 cards are jacks.

Therefore,

$P(E)$=$\dfrac{8}{{52}} = \dfrac{2}{{13}}$i.e.., the probability of drawing a card which is either king or queen.

$P(F) = \dfrac{8}{{52}} = \dfrac{2}{{13}}$i.e.., the probability of drawing a card which is either queen or jack.

There are exactly 4 cards which are “king or queen” and “queen or jack”. i.e.., drawing only queen cards.

$P(E \cap F) = \dfrac{4}{{52}} = \dfrac{1}{{13}}$ i.e., the probability of drawing a card which is a queen from a deck of 52 cards.

Now, to prove $E$ and $F$events to be independent the following condition to be satisfied.

$P(E \cap F) = P(E).P(F) \to (1)$

So let us substitute the obtained values of$P(E)$, $P(F)$and $P(E \cap F)$in equation (1) we get,

$

\Rightarrow \dfrac{1}{{13}} = \dfrac{2}{{13}}.\dfrac{2}{{13}} \\

\Rightarrow \dfrac{1}{{13}} \ne \dfrac{4}{{169}} \\

$

Therefore, we can say the events E and F are not independent as they are not satisfying the condition of independent events.

Therefore, in two cases i.e.., (i), (ii) the events E and F are independent.

Note: If A, B are the events of a sample space S are said to be independent only if they are pairwise independent i.e., $P(A \cap B) = P(A).P(B)$.

Complete step-by-step answer:

i. Given,

E: ‘the card drawn is spade’

F: ‘the card drawn is an ace’

In a deck of 52 cards, 13 cards will be spade and 4 cards are ace and only 1 card is an ace of spades.

Therefore,

$P(E)$=$\dfrac{{13}}{{52}} = \dfrac{1}{4}$i.e.., the probability of drawing a spade from a deck of 52 cards.

$P(F) = \dfrac{4}{{52}} = \dfrac{1}{{13}}$i.e.., the probability of drawing an ace from a deck of 52 cards.

$P(E \cap F) = \dfrac{1}{{52}}$ i.e.., the probability of drawing a card which is spade as well as an ace from a deck of 52 cards.

Now, to prove $E$ and $F$events to be independent the following condition to be satisfied.

$P(E \cap F) = P(E).P(F) \to (1)$

So let us substitute the obtained values of$P(E)$, $P(F)$and $P(E \cap F)$in equation (1) we get,

$

\Rightarrow \dfrac{1}{{52}} = \dfrac{1}{4}.\dfrac{1}{{13}} \\

\Rightarrow \dfrac{1}{{52}} = \dfrac{1}{{52}}[\therefore L.H.S = R.H.S] \\

$

Therefore, we can say the events $E$ and $F$ are independent as they are satisfying the condition of independent events.

ii. Given,

$E$: ‘the card drawn is black’

$F$: ‘the card drawn is a king ’

In a deck of 52 cards, 26 cards are black and 4 cards are kings only 2cards are black as well as kings.

Therefore,

$P(E)$=$\dfrac{{26}}{{52}} = \dfrac{1}{2}$i.e.., the probability of drawing a black card from a deck of 52 cards.

$P(F) = \dfrac{4}{{52}} = \dfrac{1}{{13}}$i.e.., the probability of drawing a king from a deck of 52 cards.

$P(E \cap F) = \dfrac{2}{{52}} = \dfrac{1}{{26}}$ i.e., the probability of drawing a card which is black as well as king from a deck of 52 cards.

Now, to prove $E$ and $F$events to be independent the following condition to be satisfied.

$P(E \cap F) = P(E).P(F) \to (1)$

So let us substitute the obtained values of$P(E)$, $P(F)$and $P(E \cap F)$in equation (1) we get,

$

\Rightarrow \dfrac{1}{{26}} = \dfrac{1}{2}.\dfrac{1}{{13}} \\

\Rightarrow \dfrac{1}{{26}} = \dfrac{1}{{26}}[\therefore L.H.S = R.H.S] \\

$

Therefore, we can say the events $E$and $F$ are independent as they are satisfying the condition of independent events.

iii. Given,

$E$: ‘the card drawn is a king or queen’

$F$: ‘the card drawn is a queen or jack’

In a deck of 52 cards, 4 cards are kings, 4 cards are queens and 4 cards are jacks.

Therefore,

$P(E)$=$\dfrac{8}{{52}} = \dfrac{2}{{13}}$i.e.., the probability of drawing a card which is either king or queen.

$P(F) = \dfrac{8}{{52}} = \dfrac{2}{{13}}$i.e.., the probability of drawing a card which is either queen or jack.

There are exactly 4 cards which are “king or queen” and “queen or jack”. i.e.., drawing only queen cards.

$P(E \cap F) = \dfrac{4}{{52}} = \dfrac{1}{{13}}$ i.e., the probability of drawing a card which is a queen from a deck of 52 cards.

Now, to prove $E$ and $F$events to be independent the following condition to be satisfied.

$P(E \cap F) = P(E).P(F) \to (1)$

So let us substitute the obtained values of$P(E)$, $P(F)$and $P(E \cap F)$in equation (1) we get,

$

\Rightarrow \dfrac{1}{{13}} = \dfrac{2}{{13}}.\dfrac{2}{{13}} \\

\Rightarrow \dfrac{1}{{13}} \ne \dfrac{4}{{169}} \\

$

Therefore, we can say the events E and F are not independent as they are not satisfying the condition of independent events.

Therefore, in two cases i.e.., (i), (ii) the events E and F are independent.

Note: If A, B are the events of a sample space S are said to be independent only if they are pairwise independent i.e., $P(A \cap B) = P(A).P(B)$.

Last updated date: 03rd Oct 2023

•

Total views: 363k

•

Views today: 9.63k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Who had given the title of Mahatma to Gandhi Ji A Bal class 10 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

How many millions make a billion class 6 maths CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE