With replacement - a) What is the probability of drawing a $10,J,Q$ in that order?

Without replacement - b) What is the probability of drawing a $10,J,Q$ in the order?

With instead of - c) illustration at the very least one ace (I got $= left(frac452 ight)^3$).

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Without replacement - d) illustration at the very least one ace (I gained $= left(frac452 ight)left(frac351 ight)left(frac250 ight)$)

Having trouble through a) and b) I understand that order problem so i would have to use combinations.

Any thoughts? Thanks.


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edited might 8 "16 at 23:10
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N. F. Taussig
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asked might 8 "16 in ~ 22:51
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a) In other words, it must be the situation that you gain a 10, then a J, then a Q. Since the draws space with replacement, then the probabilities don"t change and the draws are independent native one one more (assuming the deck is shuffled after replacement). Then us have$$P(TJQ) = P(T)P(J)P(Q) = frac452cdotfrac452cdotfrac452.$$

b) Again, it should be the situation that you gain a 10, climate a J, climate a Q. Due to the fact that the draws are without replacement, climate the probabilities do change and the draws room not independent from one another. After each draw, the number of cards in deck reduces by one, and also hence$$P(TJQ) = frac452cdot frac451cdotfrac450.$$

c) No, what friend calculated is the opportunity that girlfriend got exactly 3 Aces (or really any kind of rank) in three draws through replacement.

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A direct calculation involves find the opportunity of an Ace in the very first or 2nd or third draw and using inclusion exclusion.

Alternatively, we have the right to use the complement: No Aces in three draws. This is equal to $(48/52)^3$. Hence$$P( extAt least one Ace) = 1-P( extNo Aces) = 1-left(frac4852 ight)^3.$$

d) Again, we use the complement. There space $48$ non Aces, and also in three draws us must draw a no Ace. After each draw, the number of non Aces walk down, and also so$$P( extAt least one Aces)=1- P( extNo Aces) = 1-frac4852cdotfrac4751cdotfrac4650.$$