2
$\begingroup$

I am fairly confident with most of these probability and combinatorics exercises, but would like my reasoning looked over so I can move on to more difficult questions. I am unsure if I correct with part d. Thank you for the help.

1. True or False, Justify.

(a) Flip a fair coin $3$ times, probability of $1$ Heads and $2$ Tails is $1/8$.

False, $3/8$ is the probability of $1$ Head and $2$ Tails since order doesn’t matter.

(b) In a class of 72 students, at least two will share a birthday.

Though it is very likely, you need $366$ people with at least one person having a birthday on each day to guarantee two people will share a birthday.

(c) Your neighbor has two children, at least one is a girl. The probability that both are girls is $1/3$.

True, since you do not know the order and the chances of having two boys is $0$, there is a $1/3$ chance that the other child is a girl.

(d) The number of numbers with distinct digits is $9!+8!+7!+6!+5!+4!+3!+2!+1!$.

I do not know how to approach this problem. My guess is $10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ since the digits cannot be repeating.

(e) If you draw $2$ cards from a standard deck of $52$ cards, the probability that they are both black is $(1/2)^2$.

There is a less than $25\%$ chance since drawing one black card decreases the supply. The chances are $$\frac{1}{2} \cdot \frac{25}{51} = 24.41\%$$

$\endgroup$
  • $\begingroup$ I think if a number consists of more than 1 digit, then the first digit can't be zero. $\endgroup$ – kludg Oct 15 '17 at 5:11
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Oct 15 '17 at 13:07
0
$\begingroup$

Flip a coin three times. The probability of $1$ heads and $2$ tails is $1/8$.

The statement is indeed false. Assuming the coin is fair, it would be true if we were calculating the probability of obtaining one head and two tails in that order. However, the heads could occur on the first, second, or third flip. If the coin is fair, then the probability of obtaining exactly one heads in three flips of the coin is $$P(X = 1) = \binom{3}{1}\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^2$$ Here, I have used the binomial distribution. Let $p$ be the probability that an event will occur during a given trail. Then the probability that it will not occur during a given trial is $1 - p$. The binomial probability that a given event will occur exactly $k$ times in $n$ trials is
$$P(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$

You could solve the problem more simply by listing the eight events in the sample space, observing that they are equally likely if the coin is fair, and observing that the three favorable events are $HTT, THT, TTH$.

In a class of $72$ students, at least two will share a birthday.

You are correct.

Your neighbor has two children. At least one is a girl. The probability that both are girls is $1/3$.

Assuming boys and girls are equally likely, you are correct.

The number of numbers with distinct digits is $9! + 8! + 7! + 6! + 5! + 4! + 3! + 2! + 1!$.

The problem could be more clearly stated. I will assume the numbers in question are positive integers.

Since there are only $10$ decimal digits, such a number has at most $10$ digits.

Ten-digit positive integers with distinct digits: There are $9$ ways to fill the leading digit since we cannot use $0$. Once the leading digit has been filled, we can fill the remaining nine digits in $9!$ ways. Hence, there are $$9 \cdot 9!$$ ten-digit positive integers with distinct digits.

Nine-digit positive integers with distinct digits: There are $9$ ways to fill the leading digit with a digit other than $0$. Once the leading digit has been filled, we can fill the remaining eight digits in $$9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = \frac{9!}{1!}$$ ways. Hence, there are $$9 \cdot \frac{9!}{1!}$$ nine-digit positive integers with distinct digits.

Can you continue?

If you draw two cards from a standard deck of $52$ cards, the probability that they are both black is $(1/2)^2$.

You are correct.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.