If Gambit has a deck of 54 cards (a standard deck with two jokers), what is the probability that he will pick a queen followed by a joker? (Assume no replacement)

To calculate probability of multiple independent events, we need to calculate the probability of each event and multiply them together.

Recall that probability of an event can be found via the following:

$\displaystyle P=\frac{\#of desired outcomes}{Total\# of outcomes}$

So the probability of selecting a queen in the first pick is:

$\displaystyle P_Q=\frac{4}{54}=\frac{2}{27}$

And the probability of picking a joker in the second pick is:

$\displaystyle P_{Joker}=\frac{2}{53}$

So the probability of a queen followed by a joker is:

$\displaystyle \frac{2}{27}*\frac{2}{53}=\frac{4}{1431}$

Making our answer: $\displaystyle \frac{4}{1431}{}$

In order to find the probability of these both happening, you need to find the probability of each event and then multiply them together. 

The probability of picking an in-person tutoring student is $\displaystyle \frac{63}{150}$

The probability of picking a student who needs math tutoring is $\displaystyle \frac{43}{150}$

To find the probability of both of these events happening you multiply: $\displaystyle \frac{63}{150}\times\frac{43}{150}=\frac{2709}{22500}$ or you can also do $\displaystyle .42*.28\bar{6}=.1204$

We have four suits which are diamond, club, heart spade. Out of the four suits, diamond and heart are red. Therefore we will have two aces. Our probability will be what we want divided by the possibilities we have.

$\displaystyle \frac{2}{52}=\frac{1}{26}$

Anytime we see the word either, this means we are adding probabilities.

We are looking for a blue or black ball.

Our total number of blue and black balls is fourteen.

The total number of balls is twenty.

So our probability is: 

$\displaystyle \frac{14}{20}=\frac{7}{10}$

Anytime we see the word and, we need to meet conditions. Therefore we need to multiply the probability.

The probability of getting a blue ball is $\displaystyle \frac{8}{20}$ or $\displaystyle \frac{2}{5}$.

Next, the probability of getting a black ball is $\displaystyle \frac{6}{19}$ because it says without replacement, this means by drawing a black ball, we only have $\displaystyle 19$ balls left.

Now, we multiply the fractions.

$\displaystyle \frac{2}{5}*\frac{6}{19}=\frac{12}{95}$

Step 1: Find out how many numbers I can roll on a die.

I can roll $\displaystyle 6$ numbers: $\displaystyle 1,2,3,4,5,6$.

Step 2: Find the probability of rolling a $\displaystyle 1$..

Each number on a die has a probability of $\displaystyle 1/6$ of being rolled.

A die has six faces with one number on each face.

The probability of rolling a six is:  $\displaystyle \frac{1}{6}$

There are four suits in a standard 52 card deck.  There are 13 cards of a particular suit.   A spade consists of 13 of the 52 cards.  Write the probability of drawing a spade.

$\displaystyle \frac{13}{52} = \frac{1}{4}$

Multiply the probabilities.

$\displaystyle \frac{1}{6}\times \frac{1}{4} = \frac{1}{24}$

The answer is:  $\displaystyle \frac{1}{24}$

Each coin toss is independent of one another.

The probability of landing on heads or tails is an even $\displaystyle 50\%$, or $\displaystyle \frac{1}{2}$, per trial.

For the next two coins rolling heads, Bob's probability of landing both heads is:

$\displaystyle (\frac{1}{2})^2 = \frac{1}{4}$

The answer is:  $\displaystyle \frac{1}{4}$

Each toss of the fair coin will result in either heads or tails.  

Write out all the combinations of the results for each of the coin tosses.  

Let:

$\displaystyle \textup{H=Heads}$

$\displaystyle \textup{T=Tails}$

The combinations are:

Three heads:  $\displaystyle \textup{HHH}$

Two heads:  $\displaystyle \textup{HHT, HTH, THH}$

One head:  $\displaystyle \textup{HTT, THT, TTH}$

No head:  $\displaystyle \textup{TTT}$

Out of the eight possible combinations, only one of the possibilities have no heads.

The answer is:  $\displaystyle \frac{1}{8}$

The first die can either be a five or a one. Assuming a standard cubic dice, this would be a $\displaystyle \frac{2}{6}$ or $\displaystyle \frac{1}{3}$ chance.

However, the second cube would need to be the remaining number. This is only a $\displaystyle \frac{1}{6}$ probability .

Multiplying the two together:

$\displaystyle \frac{1}{3}*\frac{1}{6}=\frac{1}{18}$

This can be more easily visualized by drawing a map off all possibilities