probability help please

[mathdumb]

What is the probability of drawing a queen from a 52 card deck or rolling a 1 on a die?

 

[tkhunny]

1) Change your screen name. You cannot expect to learn anything with an attitude like that.

2) Are your events independent?

3) Catalogue: There are only four posibilities:
3a) Queen and One
3b) Not Queen and Not One
3c) Not Queen and One
3d) Queen and Not One

4) Understanding: Pr(Queen or One) = 1 - Pr(Queen and One)

5) Draw a 2x2 box.
5a) Label the columns Queen and Not Queen
5b) Label the rows One and Not One
5c) Upper left box is Pr(Queen and One). Fill in this value
5d) Upper right is Pr(Not Queen and One). Fill in this value
5ef) Continue.

6) Note: When it is small, you can just count it. However, try to learn how to track it without physically enumerating everything. Only very small problems can be handled in this way.

 

[soroban]

Hello, mathdumb!

What is the probability of drawing a Queen from a 52-card deck or rolling a "1" on a die?

\(\displaystyle \text{You are expected to know this formula: }\;P(A \vee B) \;=\; P(A) + P(B) - P(A \wedge B)\)


\(\displaystyle \text{We know that: }\;\begin{Bmatrix}P("Q") &=& \frac{4}{52} &=& \frac{1}{13} \\ \\[-3mm] P("1") &=& \frac{1}{6} \end{Bmatrix}\)

\(\displaystyle \text{Since "Q" and "1" are }indepedent\text{ events: }\;P("Q" \wedge "1") \:=\:\frac{1}{13}\cdot\frac{1}{6} \:=\:\frac{1}{78}\)


\(\displaystyle \text{Therefore: }\;P("Q" \vee "1") \;=\;P("Q") + P("1") - P("Q" \wedge "1")\)

. . . . . . . . . . . . . . . . . .\(\displaystyle =\; \frac{1}{13} + \frac{1}{6} - \frac{1}{78} \;=\;\frac{18}{78} \;=\;\frac{9}{39}\)