# Thread: 2 Questions----I need clarification.

1. ## 2 Questions----I need clarification.

The following are 2 practice questions for my math class. There will be similar questions on my test so I need to know how to do them correctly. Any help would be great, thanks!

1) Following are 3 boxes containing letters: [tex] [AND] [HISTORY].

a) From box 1, three letters are drawn one-by-one without replacement and recorded in that order. What is the probability that the outcome is HAT?

-I did (1/4)(1/3)(1/2) = 1/24 Is this correct?

b) If a box is chosen at random and then a letter is drawn at random from the box, what is the probability that the outcome is A?

-I'm thinking along the lines of (2/3)(1/4)(1/3), but am unsure...

2) Box 1: [2 black balls & 3 white balls] Box 2: [4 black balls & 2 white balls] A ball is drawn at random from box 1, then at random from box 2. The colors of the balls are recorded in order. The boxes look like this:
1:[BBWWW] 2:[BBBBWW] Find:

a) The probability of 2 whites.

-I did (3/5)(2/6) = 1/5 Is this correct?

b) The probability of at least one black. (I'm stumped)

c) The probability of at most one black. (I'm stumped again)

d) The probability of a black followed by a white, or a white followed by a black.

-I did (2/5)(2/6) = 2/15 or (3/5)(4/6) = 2/5 Is this correct?

2. Hello, mdenham2!

1) There are 3 boxes containing letters: . $[M,A,T,H]\;\; [A,N,D]\;\; [H,I,S,T,O,R,Y].$

a) From box 1, three letters are drawn one-by-one without replacement.
What is the probability that the outcome is $HAT$?

I did: . $\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2} \,=\,\frac{1}{24}$ . . . Is this correct? . . Yes!

b) If a box is chosen at random and then a letter is drawn at random from the box,
what is the probability that the outcome is $A$?
The box is chosen at random: . $P(\text{box 1}) \,=\,P(\text{box 2}) \,=\,P(\text{box 3}) \,=\,\frac{1}{3}$

From box 1, $P(A) = \frac{1}{4}$
. . Hence: . $P(\text{box 1 }\wedge \text{ A}) \:=\:\frac{1}{3}\cdot\frac{1}{4} \:=\:\frac{1}{12}$

From box 2, $P(A) = \frac{1}{3}$
.Hence: . $P(\text{box 2 }\wedge\text{ A}) \:=\:\frac{1}{3}\cdot\frac{1}{3} \:=\:\frac{1}{9}$

From box 3, $P(A) = 0$
. . Hence: . $P(\text{box 3 }\wedge\text{ A}) \:=\:\frac{1}{3}\cdot0 \:=\:0$

Therefore: . $P(A) \;=\;\frac{1}{12} + \frac{1}{9} + 0 \;=\;\frac{7}{36}$

2) Box 1: [2 black balls & 3 white balls]
. . Box 2: [4 black balls & 2 white balls]

A ball is drawn at random from box 1, then at random from box 2.

a) The probability of 2 whites.

I did: . $\frac{3}{5}\cdot\frac{2}{6} \,=\, \frac{1}{5}$ . . . Is this correct? . Yes!

b) The probability of at least one black.
The opposite of "at least one black" is "no blacks" (both white).

You found: $P(\text{2 whites}) = \frac{1}{5}$ .in part (a).

Therefore: . $P(\text{at least one black}) \;=\;1-\frac{1}{5} \;=\;\frac{4}{5}$

c) The probability of at most one black.
The opposite of "at most one black" is "both black".

$P(\text{both black}) \:=\:\frac{2}{5}\cdot\frac{4}{6} \:=\:\frac{4}{15}$

Therefore: . $P(\text{at most one black}) \;=\;1-\frac{4}{15} \;=\;\frac{11}{15}$

d) The probability of a black followed by a white, or a white followed by a black.

I did : . $\frac{2}{5}\cdot\frac{2}{6} \,=\,\frac{2}{15}\:\text{ or }\:\frac{3}{5}\cdot\frac{4}{6} \,=\,\frac{2}{5}$ . . . Is this correct? . Um, not quite ...

With an "or" statement, we add the probabilties.

Therefore: . $P(\text{BW }\vee\text{ WB}) \;=\;\frac{2}{15} + \frac{2}{5} \;=\;\frac{8}{15}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's an alternate approach to part (d).

We want: . $P(\text{opposite colors})$

In part (a), we found that: . $P(\text{2 whites}) \,=\,\frac{1}{5}$

In part (c), we found that: . $P(\text{2 blacks}) \,=\,\frac{4}{15}$

. . Hence: . $P(\text{same color}) \:=\:\frac{1}{5} + \frac{4}{15} \:=\:\frac{7}{15}$

Therefore: . $P(\text{opposite colors}) \;=\;1 - \frac{7}{15} \;=\;\frac{8}{15}$

3. ## Thank you!

Thank you so much, I couldn't have asked for better explanations!

4. Thanks!!!