Archive for the ‘probability’ Category

Problem 1166

If the probability is 0.81 that a candidate wins the election, what is the probability that he loss?

 

Solution:-

 

Suppose that we divide a final sample space into two sunset E and E’ such that E∩E’ = ø, that is they are mutually exclusive, and EUE’ = S. Then E’ called the complement of E relative to S. Thus, E’ contains all the elements of S that are not in E.

Because the events are mutually exclusive, P(E’) = 1- P(E).

Let E = the candidate wins and let E’ = the candidate, use P(E’) = 1 –P(E) to find the probability that the candidate loses the election.

Substitute P(E) into the equation below.

P(E’) = 1-0.81

Calculate P(E’)

P(E’) = 0.19

Therefore, the probability that the candidate loses the election is 0.19.

 

Problem 1165

A single card is drawn from a standard 52-cards deck. Let R be the event that the card drawn is red, and let F be the event that the card drawn is a face card. Find the indicated probability.
P(R’ U F)
Solution:-

 

P(R’ U F) = \frac{8}{13}

 

 

Problem 1164

Is the selection a permutation, a combination, or neither?

A group of 10 friends sits in the same row in a movie theater.

 

Solution:-

 

Use the definition of permutation and combination to determine the type of selection mode.

A permutation of a set of n distinct objects is an arrangement of the objects in a specific order without repetition.

A combination of a set of n distinct objects in an arrangement of the objects without repetition where order is irrelevant.

In the selection process of the problem, the order in which the friends sit is important.

There will not be any repetition since a person cannot sir in multiple seats at the same time.

Use the fact that the selection process has no repetition and order is relevant to determine the type of selection.

The selection is a permutation.

 

Problem 1163

An experiment consists of rolling two fair dice and adding the dots on the two sides facing up.

Find the probability of the sum of the dots indicated.

A sum loss then 7

 

Solution:-

 

The probability of getting a sum less than 7 is \frac{5}{12}

 

Problem 1151

A health inspector must visit 4 of restaurants on Monday. In how many can she pick a first, second, third, and fourth restaurant visit?

 

Solution:-

 

The number of ways the inspector can pick the first restaurant is 10.

The number of ways the inspector can pick the second restaurant is 9.

Continuing in this manner, you can see that this a permutation of 10 things taken 4 at a time.

P(10,4) = \frac{10!}{(10-4)!}=\frac{10!}{6!} = 5040

She can visit the restaurants in 5040 ways.

 

Problem 1150

A man has 9 shirt and 4 ties. How many different shirt and tie arrangements can be wear?

 

Solution:-

 

Multiplication Principle of Counting.

If a  take consists of a sequence of choices in which there are p selections for the first choice, q  selection for the  second choice, r selection for the third choice, and so no, then task of making these selections can be done in

P*q*r*…

different ways.

There are only two choice to be made. First the man must select a shirt to wear, and then he has to select a tie.

There are 9 ways for him to select a shirt.

There are 4 ways for him to select a tie.

So, by the Multiplication Principle of Counting, there are 9 * 4 or 36 different shirt and tie arrangements from which to choose.

 

 

Problem 1149

Pizza House offers 4 different salads, 7 different kinds of pizza, and 5 different desserts. How many different there course meals can be ordered?

 

Solution:-

 

For the event of choosing a salad, the number of possible outcomes is 4.

For the event of choosing a pizza, the number of possible outcomes is 7.

For the event of choosing a dessert, the number of possible outcomes is 5.

Applying the multiplication Principle you have 4 * 7*5 = 140.

 

Problem 1148

Determine whether the following pair of events are dependent or independent.

Event E is the event that a resident of Pennsylvania lives in Pittsburgh, and F is the event that a resident of Pennsylvania lives in either Pittsburgh or Philadelphia.

 

Solution:-

 

Two events, E and F, are dependent if the outcome of event E influences the outcome of event F, and vice versa. Event E and F are independent if the outcome of E does not affect the outcome of event F, and vice versa.

Notice that the fact that a resident of Pennsylvania lives in Pittsburgh influences the fact that a resident of Pennsylvania lives in either Pittsburgh or Philadelphia, because if a resident of Pennsylvania lives in Pittsburgh, it is certain that the resident of Pennsylvania lives in Pittsburgh or Philadelphia.

Therefore, the two events are dependent.

 

Problem 1147

Determine whether the following pair of events are dependent or independent.

Two teams from different sport play a game. A is the event that the first team wins, and B is the event the second team wins.

 

Solution:-

 

Two events, E and F, are dependent if the outcome of event E influences the outcome of event F, and vice versa. Events E and F are independent if the outcome of E does not affect the outcome of event F, and vice versa.

The fact that the first team wins does not influence the result of the second team, and vice versa.

Therefore, the two events are independent.

 

 

 

Problem 1146

Two cards are drawn without replacement from an ordinary deck. Find the probability that two aces are drawn.

 

Solution:-

 

Notice that the problem statement can be written as P(E∩ F), where E and F represent the two events. The product rule of probability says that P(E ∩ F) = P(E).P(F|E).

First find P(E), the probability that the first card drawn is an ace.

P(E) = \frac{4}{52}=\frac{1}{13}

To find P(F|E) in this case, find the probability of drawing ace for the second card. Notice that there are 51 cards left in the deck an ace is drawn.

This partial deck of 51 cards is the reduced sample space used to calculate the probability when the second card is drawn.

After the first card is drawn from the deck, there are 3 aces left in the deck.

Thus, P(F|E), the probability that the second card chosen is an ace given the first card is an ace is \frac{3}{51} , or \frac{1}{17}.

Finally, find P(E∩F).

P(E∩F) = P(E).P(F|E)

=\frac{1}{13}.\frac{1}{17}

Therefore, the probability that two aces are drawn is \frac{1}{221}.