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