Respuesta :
Answer:
(a)No. The probability of drawing a specific second card depends on the identity of the first card.
(b)4/663
(c) 4/663
(d) 8/663
Step-by-step explanation:
(a)The events are not independent because we are drawing cards without replacement and the probability of drawing a specific second card depends on the identity of the first card.
(b) P(ace on 1st card and jack on 2nd).
[tex]P$(Ace on 1st card) =\dfrac{4}{52}\\ P$(Jack on 2nd card)=\dfrac{4}{51}\\\\$Therefore:\\P(ace on 1st card and jack on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}[/tex]
(c)P(jack on 1st card and ace on 2nd)
[tex]P$(Jack on 1st card) =\dfrac{4}{52}\\ P$(Ace on 2nd card)=\dfrac{4}{51}\\\\$Therefore:\\P(jack on 1st card and ace on 2nd) =\dfrac{4}{52}\times \dfrac{4}{51}\\=\dfrac{4}{663}[/tex]
(d)Probability of drawing an ace and a jack in either order.
We can either draw an ace first, jack second or jack first, ace second.
Therefore:
P(drawing an ace and a jack in either order) =P(AJ)+(JA)
From parts (b) and (c) above:
[tex]P$(jack on 1st card and ace on 2nd) =\dfrac{4}{663}\\P$(ace on 1st card and jack on 2nd) =\dfrac{4}{663}\\$Therefore:\\P(drawing an ace and a jack in either order)=\dfrac{4}{663}+\dfrac{4}{663}\\=\dfrac{8}{663}[/tex]