Part (a)
There are 3 face cards (jack, queen, king) in any given suit. We have 4 suits. That gives 3*4 = 12 face cards total.
12/52 = 3/13 is the probability of getting a face card on any random selection. This probably does <u>not</u> change for any future selection because we put the current card back. In other words, the deck has not changed. Contrast this with part (b) where the probabilities will change since the card counts decrease each time (since we don't put the card back).
The probability of getting 3 face cards in a row, with replacement, is:
(3/13)*(3/13)*(3/13) = 0.012289 approximately
This converts to the percentage 1.2289% after moving the decimal point two spots to the right. This is the same as multiplying by 100.
Lastly, round that to one decimal place to get 1.2%
<h3>
Answer: 1.2</h3>
The percent sign is already taken care of, so you'll just input the number only.
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Part (b)
Like mentioned earlier, the probability of a face card is 12/52
I won't reduce that fraction, as you'll see why below.
After that first card is <u>not</u> put back, the 12 decreases to 12-1 = 11 and we have 52-1 = 51 cards left over.
11/51 is the probability of getting another face card, where we have no replacement going on.
Then 10/50 is the probability of getting a third face card. Again, with no replacement.
Therefore,
(12/52)*(11/51)*(10/50) = 0.009955
is the approximate probability of getting 3 face cards in a row with no replacement. Notice the count down 12,11,10 in the numerators. There's also the countdown 52, 51, 50 in the denominators. This nice pattern only happens if we don't reduce the reducible fractions. This is why I didn't reduce.
Anyways 0.009955 converts to 0.9955% which rounds to 1.0% or simply 1%
<h3>Answer: 1</h3>
where 
is the length of one side. (This comes from the general rectangle formula 
with 
).
