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Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. 1/9
B. 2/9
C. 3/9
D. 4/9
E. 5/9


We only need to look at the 4th digit (last card) to judge if the number is even. Although the cards are drawn one by one this doesn't affect the distribution of each card in the grand scheme, the fourth card will still have a 1/9 chance of each digit. Thus it is a 4/9 chance of it being even.

Ans: D
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The total number of 4 digits that may be formed is 9*8*7*6 = 3024.

For the 4 digit number to be even, last digit must be even. Here since 4 digit number is to be formed and there are 4 even numbers. so even if we have the first 3 digit even, we can spare an even number for the 4th digit.
Overall, the choice of first 3 digit doesn't hamper the chance of last digit being even or odd.

So we can pick the last digit in 4 ways
and the first 3 digit in 8*7*6 ways.
The total number of such even 4 digit number = 1344

Hence the probability of the 4 digit number being even = (8*7*6*4)/(9*8*7*6) = 4/9.

Short Method.
Since the choice of first 3 digit doesn't hamper the chance of last digit being even or odd. We can simply say that the chances of the number being even is 4/9 (since there are 4 even numbers from 1 to 9.)
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Hi Bunuel
Please help me understand why the probability does not change. Imagine that the first three cards are 2 even cards, you have only 2 even cards out of 6. Or for example imagine that the first 3 picked card are 1 even and two odds, so you have 3 out of 6... Please helpe me understand why thid does not apply to the exercise. Thanks in advance!
Bunuel
Official Solution:

Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)


The four-digit number will be even if the units digit is even. So, the question basically asks about the probability that 4th card picked will be even. There are 4 even cards from 9, so the probability that 4th card will be even is \(\frac{4}{9}\).

To elaborate more: the initial probability of drawing an even card is \(\frac{4}{9}\). Without knowing the other results, the probability of drawing an even card will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).


Answer: D
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Javmex
Hi Bunuel
Please help me understand why the probability does not change. Imagine that the first three cards are 2 even cards, you have only 2 even cards out of 6. Or for example imagine that the first 3 picked card are 1 even and two odds, so you have 3 out of 6... Please helpe me understand why thid does not apply to the exercise. Thanks in advance!
Bunuel
Official Solution:

Nine cards numbered from 1 to 9 are placed in an empty bowl. Four cards are drawn from the bowl one-by-one without replacement to form a four-digit number (first card gives the thousands digit, second card gives the hundreds digit, third card gives the tens digit and fourth card gives the units digit). What it the probability that the four-digit number is even?

A. \(\frac{1}{9}\)
B. \(\frac{2}{9}\)
C. \(\frac{3}{9}\)
D. \(\frac{4}{9}\)
E. \(\frac{5}{9}\)


The four-digit number will be even if the units digit is even. So, the question basically asks about the probability that 4th card picked will be even. There are 4 even cards from 9, so the probability that 4th card will be even is \(\frac{4}{9}\).

To elaborate more: the initial probability of drawing an even card is \(\frac{4}{9}\). Without knowing the other results, the probability of drawing an even card will not change for ANY successive drawing: second, third, fourth... There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).


Answer: D

The point is that we don't know what the first three cards were, so without that information, nothing changes when calculating the probability for the fourth card.

Consider this: there are 8 cards, 5 spades, and 3 hearts. What is the probability that the first card you pick will be a spade? Clearly, it’s 5/8.

Now, suppose I randomly remove three cards without telling you which cards were thrown away. What is the probability now that you pick a spade out of the 5 remaining cards? Even though the sample has been reduced, the probability still remains 5/8. Why should it change?

Another scenario: using the same 8 cards (5 spades and 3 hearts), imagine you are allowed to pick one card, but only from a randomly selected subset of 5 cards. What is the probability of picking a spade? Again, it remains 5/8.

In all these cases, the probability does not change. If it were to increase or decrease, it would imply that the probability of picking a heart would inversely change. But there is no reason for this to happen, as the process remains random, and we lack information about the removed or previously picked cards.

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