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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
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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., $P(A \cap B) = P(A).P(B)$.
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