Answer

Verified

473.1k+ 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., $P(A \cap B) = P(A).P(B)$.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

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

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

One cusec is equal to how many liters class 8 maths CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

The mountain range which stretches from Gujarat in class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths