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Mathematics

1 answer:

Alja [10]2 years ago

5 0

Answer:

false

Step-by-step explanation:

it's false negative

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The following question involves a standard deck of 52 playing cards. In such a deck of cards there are four suits of 13 cards ea

olasank [31]

Answer:

(a)No. The probability of drawing a specific second card depends on the identity of the first card.

(b)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).

$P$(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}$

(c)P(jack on 1st card and ace on 2nd)

$P$(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}$

(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:

$P$(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}$