# Math Help - Coin Flip Questions, Shooting Question, And Rock Concert?

1. ## Coin Flip Questions, Shooting Question, And Rock Concert?

#1 A fair coin is flipped 30 times. What is the probability the coin will show heads less than 17 times?

My work: 1 P(NOT 17) = 1 - (30C13)(.5)^30

#2 Tom has a 1/2 of shooting the target, and Kate has a 1/3 chance of shooting the target. What is the probability that atleast one of them will shoot the target? What is the probability that both of them will shoot the target? (This is independent)

#3 A rock concert is attended by 1300 students of whom 760 are girls and 540 are boys. If 4 of these students are randomly chosen to receive backstage passes, what is the probability that at most 3 of the chosen will be boys?

My work:
1 - P(OB, 3G)
1-(1300C3)(760/1300)^3

2. Hello, AlphaRock!

1) A fair coin is flipped 30 times.
What is the probability the coin will show heads less than 17 times?
This is an awful problem!

We want the probability that the coin will show Heads:
. . 0 times or 1 time or 2 times or 3 times or . . . 16 times.

So we must find:

. . $\begin{array}{ccc}P(\text{0 Heads}) &=& (\frac{1}{2})^{30} \\
P(\text{2 Heads}) &=& (_{30}C_2)(\frac{1}{2})^{30} \\ \vdots && \vdots \\

and add them up . . .

2) Tom has a 1/2 chance of hitting a target;
Kate has a 1/3 chance of hitting the target.

(a) What is the probability that at least one of them will hit the target?
(b) What is the probability that both of them will hit the target? (This is independent)
(a) The opposite of "at least one hits the target" is "both miss the target."

. . $P(\text{Tom misses}) = \frac{1}{2},\;P(\text{Kate misses}) = \tfrac{2}{3}$

. . $P(\text{both miss}) \:=\:\left(\frac{1}{2}\right)\left(\frac{2}{3}\rig ht) \:=\:\frac{1}{3}$

. . Therefore: . $P(\text{at least one hit}) \;=\;1 - \frac{1}{3} \;=\;\frac{2}{3}$

(b) . $P(\text{both hit}) \;=\;\left(\frac{1}{2}\right)\left(\frac{1}{3}\rig ht) \;=\;\frac{1}{6}$

3) A rock concert is attended by 1300 students: 760 girls. 540.
If 4 of these students are randomly chosen to receive backstage passes,
what is the probability that at most 3 of the chosen will be boys?

The opposite of "at most 3 boys" is "4 boys."

There are: . $_{1300}C_4$ possible outcomes.

There are: . $_{540}C_4$ ways that 4 boys can be chosen.

Hence, there are: . $(_{1300}C_4) - (_{540}C_4)$ ways that at most 3 boys are chosen.

Therefore: . $P(\text{at most 3 boys}) \;=\;\frac{(_{1300}C_4) - (_{540}C_4)}{_{1300}C_4}$

3. OMG! WOW!!! Soroban!! You cleared everything up for me! Everything so much easier now!!

Thanks!

4. Originally Posted by Soroban
Hello, AlphaRock!

This is an awful problem!

We want the probability that the coin will show Heads:
. . 0 times or 1 time or 2 times or 3 times or . . . 16 times.

So we must find:

. . $\begin{array}{ccc}P(\text{0 Heads}) &=& (\frac{1}{2})^{30} \\
P(\text{2 Heads}) &=& (_{30}C_2)(\frac{1}{2})^{30} \\ \vdots && \vdots \\

and add them up . . .

[snip]

The amount of computation can be reduced considerably with a little ingenuity.

Let H be the number of heads, which has a Binomial(n=30, p=.5) distribution. We want to find $\Pr(H \leq 16)$. By the symmetry of the Binomial distribution for p = .5, we know that

$\Pr(H \leq 14) = \Pr(H \geq 16)$.

Then since

$\Pr(H \leq 14) + \Pr(H = 15) + \Pr(H \geq 16) = 1$,

$2 \Pr(H \leq 14) + \Pr(H = 15) = 1$.

So

$\Pr(H \leq 14) = (1/2) \; (1 - \Pr(H = 15))$.

And

$\Pr(H \leq 16) = \Pr(H \leq 14) + \Pr(H = 15) + \Pr(H = 16)$,

so

$\Pr(H \leq 16) = (1/2) \; (1 + \Pr(H =15)) + \Pr(H = 16)$

which requires only two evaluations of the Binomial pdf.

(Of course, if you have calculator which gives you the cumulative distribution automatically, you may not care.)

5. Originally Posted by awkward
The amount of computation can be reduced considerably with a little ingenuity.

Let H be the number of heads, which has a Binomial(n=30, p=.5) distribution. We want to find $\Pr(H \leq 16)$. By the symmetry of the Binomial distribution for p = .5, we know that

$\Pr(H \leq 14) = \Pr(H \geq 16)$.

Then since

$\Pr(H \leq 14) + \Pr(H = 15) + \Pr(H \geq 16) = 1$,

$2 \Pr(H \leq 14) + \Pr(H = 15) = 1$.

So

$\Pr(H \leq 14) = (1/2) \; (1 - \Pr(H = 15))$.

And

$\Pr(H \leq 16) = \Pr(H \leq 14) + \Pr(H = 15) + \Pr(H = 16)$,

so

$\Pr(H \leq 16) = (1/2) \; (1 + \Pr(H =15)) + \Pr(H = 16)$

which requires only two evaluations of the Binomial pdf.

(Of course, if you have calculator which gives you the cumulative distribution automatically, you may not care.)
Thanks for showing a quicker way, awkward.

I'm having troubles understanding how to get the probability using this way. Can somebody help clear things up (by using the C)?

6. Originally Posted by AlphaRock
Thanks for showing a quicker way, awkward.

I'm having troubles understanding how to get the probability using this way. Can somebody help clear things up (by using the C)?
For a Binomial(n = 30, p = 0.5) distribution,

$\Pr(H = 15) = \binom{30}{15} (1/2)^{30}$

and

$\Pr(H = 16) = \binom{30}{16} (1/2)^{30}$.

Is that what you mean?

7. Originally Posted by AlphaRock
#1 A fair coin is flipped 30 times. What is the probability the coin will show heads less than 17 times?

My work: 1 P(NOT 17) = 1 - (30C13)(.5)^30
This is the probability that the coin shows heads anynumber of times except 17, either less than or larger than 17.

#2 Tom has a 1/2 of shooting the target, and Kate has a 1/3 chance of shooting the target. What is the probability that atleast one of them will shoot the target? What is the probability that both of them will shoot the target? (This is independent)

The probabilty that both will miss is (1/2)(2/3)= 1/3 so the probability that at least one will hit is 1- 1/3= 2/3.
The probability that both will hit is (1/2)(1/3)= 1/6.

#3 A rock concert is attended by 1300 students of whom 760 are girls and 540 are boys. If 4 of these students are randomly chosen to receive backstage passes, what is the probability that at most 3 of the chosen will be boys?
The "opposite" of "at most 3 will be boys" is "all 4 will be boys". There are 1300 students, of whom 540 are boys. The probability that the first student chosen is a boy is 540/1300. Assuming that a boy is chosen, there are 1299 students left of whom 539 are boys. The probability the second student chosen is a boy is 539/1299. Similarly, the probability that the third student chosen is a boy is 538/1298 and the probability that the fourth student is a boy is 537/1297. The probability that all four student chosen are boys is (540/1300)(539/1299)(538)/1298)(537/1297). The probability that does NOT happen, the probability that at most 3 boys are chosen, is 1- (540/1300)(539/1299)(538)/1298)(537/1297).

My work:
1 - P(OB, 3G)
0 boys 3 girls? But that would be only 3 students chosen.

1-(1300C3)(760/1300)^3