# Intro to probability questions

• January 27th 2010, 03:42 AM
rozekruez
Intro to probability questions
If the probability of the Leafs defeating Senators in a hockey game is 3/7, what is the probability that the Leafs will win two consecutive games against the Senators?

A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
• January 27th 2010, 03:48 AM
e^(i*pi)
Quote:

Originally Posted by rozekruez
If the probability of the Leafs defeating Senators in a hockey game is 3/7, what is the probability that the Leafs will win two consecutive games against the Senators?

Assuming the two events are independent then the second match has the same chance as the first. To find AND probabilities you should multiply each chance:

Spoiler:
$\left(\frac{3}{7}\right)^2 = \frac{9}{49}$

Quote:

A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
As the outcomes are mutually exclusive we can say that

$P(R) + P(Y) + P(R,Y) = 1$

In this case it is consider the chance of not getting a yellow block and take it from 1.

Spoiler:
$P(Y or R,Y) = 1 - P(R) = 1 - \frac{1}{3} = \frac{2}{3}$

Multiply the outcome by 24 to get the total number of blocks which are not red/which have yellow on them.

Spoiler:
$\frac{2}{3} \times 24 = 16$
• January 27th 2010, 03:55 AM
$P(Red)=\frac{1}{3}=\frac{R}{24}$
$P(Red\ and\ Yellow)=\frac{1}{12}=\frac{M}{24}$