Help with question- stats 1

How many different 4-digit numbers can be formed using the digits 1,2 and 3 when each digit may be included at most twice?
This question is worth 5 marks.

Also: A random variable X has the distribution B(5,1/4).
Two values of X are chosen at random, find the probability that their sum is less than 2 [4 marks]

(edited 7 years ago)

Reply 1

Zacken

22

Original post by marmbite

Also: A random variable X has the distribution B(5,1/4).
Two values of X are chosen at random, find the probability that their sum is less than 2 [4 marks]

If X is distributed with 5 trials and probability of success 1/4.

Then X1 + X2 is distributed with 10 trials and probability of success. Can you see why this is true?So let's call this new distribution

Y \sim B(10, 0.25)

.

You want

\mathbb{P}(Y < 2) = \mathbb{P}(Y \leq 1)

which is an easy table lookup.

Reply 2

Zacken

22

Original post by marmbite

Also: A random variable X has the distribution B(5,1/4).
Two values of X are chosen at random, find the probability that their sum is less than 2 [4 marks]

Another method is (I like this question!), you pick two numbers X1 and X2, for their sum to be less than 2, the only possibilities are:

X1 = 0, X2 = 0
X1 = 0, X2 = 1
X1 = 1, X2 = 0.

So:

P(X1 = 0) * P(X2 = 0) + P(X1 = 0) * P(X2 = 1) + P(X1 = 1) * P(X2 = 0)

Where you can compute each individual probability using the binomial pdf. (If you're really sneaky, you'll notice that the last two are actually the same so you can just compute P(X1 = 0, X2=1) and then multiply it by 2).

Reply 3

Zacken

22

Original post by marmbite

How many different 4-digit numbers can be formed using the digits 1,2 and 3 when each digit may be included at most twice?
This question is worth 5 marks.

So, basically, what we're doing is splitting it into different cases.

Case (1) only 1 number repeats.

So, if we write the placeholders as _ _ _ _ (e.g: 1 1 2 3)

then we have 4! ways to re-arrange those 4 numbers but 1 repeats itself, so we have 4! / (2!) ways to re-arrange those numbers. However, that was with us assuming that 1 repeats itself. We could have had 2 repeat itself and 3 repeat itself, so that gives us 3 more combinations. So 3 * 4! / 2! is the # of ways to re-arrange the numbers with only 1 number repeating itself.

Case (2): 2 numbers repeat.

So, if we write the placeholders as _ _ _ _ (e.g: 1 1 2 2)

then we have 4! ways to re-arrange those 4 numbers but 1 and 2 repeats itself so we have 4! / (2! 2!) ways to re-arrange those numbers. However that was with us assuming that 1 and 2 repeats itself. It could have been 1 and 3 or 2 and 3 repeating itself. Again, another factor of 3 present. So we have 3 * 4! / (2! 2!) ways to re-arrange the numbers with 2 numbers repeating themselves.

Add them up to get the total number of combinations.

Reply 4

marmbite

OP

9

Original post by Zacken

So, basically, what we're doing is splitting it into different cases.

Case (1) only 1 number repeats.

So, if we write the placeholders as _ _ _ _ (e.g: 1 1 2 3)

then we have 4! ways to re-arrange those 4 numbers but 1 repeats itself, so we have 4! / (2!) ways to re-arrange those numbers. However, that was with us assuming that 1 repeats itself. We could have had 2 repeat itself and 3 repeat itself, so that gives us 3 more combinations. So 3 * 4! / 2! is the # of ways to re-arrange the numbers with only 1 number repeating itself.

Case (2): 2 numbers repeat.

So, if we write the placeholders as _ _ _ _ (e.g: 1 1 2 2)

then we have 4! ways to re-arrange those 4 numbers but 1 and 2 repeats itself so we have 4! / (2! 2!) ways to re-arrange those numbers. However that was with us assuming that 1 and 2 repeats itself. It could have been 1 and 3 or 2 and 3 repeating itself. Again, another factor of 3 present. So we have 3 * 4! / (2! 2!) ways to re-arrange the numbers with 2 numbers repeating themselves.

Add them up to get the total number of combinations.

Great explanation, thanks!

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