Given: pack of 52 cards
Formula: P(E) = \(\frac{favourable\,outcomes}{total\,possible\,outcomes}\)
since a card is drawn from a pack of 52 cards, therefore number of elementary events in the sample space is
n(S)= 52C1 = 52
(i) let E be the event of drawing a black king
n(E)=2C1 =2 (there are two black kings one of spade and other of club)
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{2}{52}\) = \(\frac{1}{26}\)
(ii) let E be the event of drawing a black card or a king
n(E) = 26C1+4C1- 2C1= 28
we are subtracting 2 from total because there are two black king which are already counted and to avoid the error of considering it twice
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{28}{52}\) = \(\frac{7}{13}\)
(iii) let E be the event of drawing a black card and a king
n(E)=2C1 =2 (there are two black kings one of spade and other of club)
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{5}{52}\) = \(\frac{1}{26}\)
(iv) let E be the event of drawing a jack, queen or king
n(E)=4C1+ 4C1+ 4C1 = 12
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{12}{52}\) = \(\frac{3}{13}\)
(v) let E be the event of drawing neither a heart nor a king now consider E’ as the event that either a heart or king appears
n(E’) = 6C1+ 4C1 - 1 = 16 (there is a heart king so it is deducted)
P(E') = \(\frac{n(E')}{n(S)} \)
P(E') = \(\frac{16}{52}\) = \(\frac{4}{13}\)
P(E) = 1- P(E’)
P(E) = 1 - \(\frac{4}{13}\) = \(\frac{9}{13}\)
(vi) let E be the event of drawing a spade or king
n(E)= 13C1 + 4C1 - 1 = 16
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{16}{52}\) = \(\frac{4}{13}\)
(vii) let E be the event of drawing neither an ace nor a king now consider E’ as the event that either an ace or king appears
n(E’) = 4C1+ 4C1 = 8
P(E') = \(\frac{n(E')}{n(S)} \)
P(E') = \(\frac{8}{52}\) = \(\frac{2}{13}\)
P(E) = 1- P(E’)
P(E) = 1 - \(\frac{2}{13}\) = \(\frac{11}{13}\)
(viii) let E be the event of drawing a diamond card
n(E)= 13C1 = 13
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{13}{52}\) = \(\frac{1}{13}\)
(ix) let E be the event of drawing not a diamond card now consider E’ as the event that diamond card appears
n(E’) = 13C1 =13
P(E') = \(\frac{n(E')}{n(S)} \)
P(E') = \(\frac{13}{52}\) = \(\frac{1}{4}\)
P(E) = 1- P(E’)
P(E) = 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)
(x) let E be the event of drawing a black card
n(E)= 26C1= 26 (spades and clubs)
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{26}{52}\) = \(\frac{1}{2}\)
(xi) let E be the event of drawing not an ace now consider E’ as the event that ace card appears
n(E’) = 4C1= 4
P(E') = \(\frac{n(E')}{n(S)} \)
P(E') = \(\frac{4}{52}\) = \(\frac{1}{13}\)
P(E) = 1- P(E’)
P(E) = 1 - \(\frac{1}{13}\) = \(\frac{12}{13}\)
(xii) let E be the event of not drawing a black card
n(E) = 26C1= 26 (red cards of hearts and diamonds)
P(E) = \(\frac{n(E)}{n(S)} \)
P(E) = \(\frac{26}{52}\) = \(\frac{1}{2}\)