Question

please! check my anwser.
Four cards drawn from standard desk of cards .Each card is replaced before the next one is drawn .Find eanh probability

Answer 1

The probability of one type of card (red, ace of diamond) is

1/52.

Answer 2

P ( 4 jacks) = (4/52) + (4/52) + (4/52) + (4/52)
P( at most 1 jack ) = ?

Answer 3

Drawing a jack is an independent event, multiply probabilities of independent events
P(4 jacks) = (4/52)(4/52)(4/52)(4/52)
P(at most 1 jack) = P(0 jacks) + P(1 Jack)
P(0 jacks) = (48/52)(48/52)(48/52)(48/52)
P(1 jack) = (4C1)(4/52)(48/52)(48/52)(48/52)

Answer 4

its similar to binomial theorem
n = 4
p = 4/52

Answer 5

for jacks means jack and jack and jack and jack. since the cards are replaced it is
\[(\frac{4}{52})^4\] oh, what dumbcow said, never mind

Answer 6

can you help me
P( exactly 3 jacks) ?

Answer 7

(4/52)^4 ?

Answer 8

yess exactly 3 jacks

Answer 9

3 jacks, one not jack
\[(\frac{4}{52})^3\times \frac{48}{52}\times 4\]

Answer 10

4= ?

Answer 11

four cards.
jack, jack, jack, not jack
jack, jack, not jack, jack
jack, not jack, jack, jack
not jack, jack, jack, jack

Answer 12

that explains why you multiply by 4

Answer 13

each one of these has the same probability, each one is
\[(\frac{4}{52})^2(\frac{48}{52})\]

Answer 14

but there are 4 disjoint possibilities, hence the multiply by 4 part