Question

From a deck of 52 playing cards all red faces are removed. If a card is drawn randomly from the remaining find the probability of
(1) getting a face card
(2)getting a black ace

Answer 1

total outcome=52-6=46
(1)favourable outcome=6
therefore probability=6/46=3/23
(2) clear out second question is not clear I think, what is ace?

Answer 2

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$\underline{\underline{\huge{\textbf{Solution:}}}}$

Given,
From a deck of 52 playing cards all red faces are removed.

Here,
Experiment - Card randomly drawn from the remaining.

Let $E_{1}$ be the Event that card drawn randomly from the remaining is a face card.

Let $E_{2}$ be the Event that card drawn randomly from the remaining is a black ace .

$\underline{\large{\textbf{Step-1:}}}$
Find Total No. of Outcomes of the Experiment.

Let S be Sample Space.

S = {(Total cards)} - {(Red face cards)}

Red Face cards are :
$\diamondsuit{K}$,
$\diamondsuit{Q}$,
$\diamondsuit{J}$,
$\heartsuit{K}$,
$\heartsuit{Q}$,
$\heartsuit{J}$.

$\therefore$ No. of Red Face Cards = 6

Total Outcomes of the Experiment n(S) = No. of ways of drawing a card from the remaining cards when red face cards are removed.

$\therefore$ n(S) = 52 - 6 = 46

$\underline{\large{\textbf{Step-2:}}}$
Find No. of Outcomes favouring occurrence if Event $E_{1}$

(Getting a Face card)

Total Face cards = 12
In which Red face cards are 6 and Black Face cards are 6.

No. of Face cards left after removing red face cards = 6.

No. of Favorable outcomes for Occurrence of Event $E_{1}$, ($n(E_{1})$) = 6

$\underline{\large{\textbf{Step-3:}}}$
Find No. of Outcomes favouring occurrence of Event $E_{2}$

(Getting a Black Ace)

Black Ace are :
$\clubsuit{A}$,
$\spadesuit{A}$

$\therefore$ No. of Black Ace = 2

No. of Favorable outcomes for Occurrence of Event $E_{2}$, ($n(E_{2})$) = 2

$\underline{\large{\textbf{Step-4:}}}$
Find Probability of occurrence of Event $E_{1}$ and $E_{2}$

We have,
$Probability\: of\: Occurrence\: of\: Event = \dfrac{No.\: of\: Favorable\: Outcomes\: for\: Event}{Total\: No.\: of\: Outcomes\: in\: Experiment}$

$P(E_{1}) = \dfrac{n(E_{1})}{n(S)}$

Substituting Values

$\implies P(E_{1}) = \dfrac{6}{46}$

$\implies P(E_{1}) =\dfrac{3}{23}$

Also,
$P(E_{2}) = \dfrac{n(E_{2})}{n(S)}$

Substituting Values

$\implies P(E_{2}) = \dfrac{2}{46}$

$\implies P(E_{2}) = \dfrac{1}{23}$

$\therefore$

Probability of getting a Face Card = $\dfrac{3}{23}$

Probability of getting a Black Ace = $\dfrac{1}{23}$

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