Posts Tagged ‘probability’

Problem 1167

A pair dice is rolled. What is the probability of getting a sum of 8?

 

Solution:-

 

To find the probability of an event, denoted P(E), you need to find the number of possible outcomes of the experiment, n(s) . This is the same as the number if elements in the sample space. You also need to find the number of ways the event can occur, denoted n(E).

Then use the formula p(E) = \frac{n(E)}{n(S)}.

The experiment in this case is the rolling of two dice.

Sum= 8({2,6},{3,5},{4,4},{5,3},{6,2})

By counting all possible outcomes it is clear that n(s) = 36.

For the event of getting a sum of 8, the number of possible outcomes is 5.

Thus n(E) = 5.

Therefore, P(E) = \frac{n(E)}{(S)} = \frac{5}{36}.

 

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 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}.

 

 

Problem 1145

Two cards are drawn without replacement from an ordinary deck. Find the probability that the second is a spade, given that the first is the ace of spades.

 

Solution:-

 

To find the conditional probability in this case, notice that the first card has been remove from the deck so the total sample space for the second card has been reduced from the original 52-card deck. Notice that after the first card is drawn from the deck, there are 51 cards left to choose from.

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 12 spades left in the deck.

Recall that if an event E is a subset of a sample space S, then the probability that event E occurs is P(E) = \frac{n()E}{n(s)}. Thus, the  probability that the second card chosen is a spade is \frac{12}{51}, or \frac{4}{17}.

Therefore, the probability that the second card is a spade given that first is the ace of spades is \frac{4}{17}.

 

Problem 1144

Two cards are drawn without replacement from an ordinary deck, find the probability that the second id a club, given that the first is not a club.

 

Solution:-

 

After the first card is draw from the deck, 51 cards are left to choose from.

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 (not a club) is drawn from the deck, 12 are left fall into the category ‘a club’.

The probability that the second is a club, given the  first is not a club is \frac{13}{51}. In lowest terms, P(a club | not a club ) = \frac{13}{51}.

 

 

Problem 1142

Is the following an example of empirical probability?

A weather forecast that predicts a 20% change of a tornado tomorrow.

 

Solution:-

 

Probabilities are empirical when it is not possible to establish  exact probabilities for the event in question. Instead, useful approximation are often found by drawing on post experience.

Let E represent the event of having a tornado tomorrow. There are many factors that affect this event, some of which are not known. It is not possible to establish an exact probability for the event.

Therefore, the given probability is an example of an empirical probability.

 

 

Problem 1141

Is the following an example of empirical probability?

The probability of getting a 6 on 4 consecutive rolls of a die.

 

Solution:-

 

Probabilities are empirical when it is not possible to establish exact probabilities for the events in question. Instead, useful approximations are often found by drawing on past experience.

Let E represent the event getting a 6 on 4 consecutive rolls of a die. Since the probability for all the outcomes for rolling a fair die are known, it is possible to calculate an exact probability for this event.

Therefore, the probability of getting a 6 on 4 consecutive rolls of a die is not an example of an empirical probability.

 

 

 

Problem 1140

A card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing an ace.

 

Solution:-

 

The probability of an event E that is a subset of a sample space S is P(E) = \frac{n(E)}{n(S)}, where n(E) is the number of outcomes in the event and n(S) is the number of outcomes in the sample space.

The sample space is the set of all possible outcomes for drawing a single card from the deck. Since there are 52 different cards in the deck, the sample space has 52 outcomes, or n(S) = 52.

A deck of 52 cards has four suits (hearts, clubs, diamonds, and spades) with 13 cards each, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, and ace. Hearts and diamonds are red, while clubs and spades are black. There are 4 ways to draw an ace.

ace  of hearts, ace of clubs, ace if diamonds, and ace of spades.

Therefore , there are n(E) = 4 outcomes in the event.

Divide n(E) = 4 by n(S) = 52 to write the probability.

P(E) = \frac{n(E)}{n(S)}

=\frac{4}{52}

=\frac{1}{13}

Therefore, the probability of drawing an ace is \frac{1}{13}.

 

 

Problem 1139

A die is tossed. What is the probability of rolling a number greater than 4?

 

Solution:-

 

The sample space is S = {1, 2, 3, 4, 5, 6}

E is the event the die rolled is greater than 4.

E = {5, 6}

P(E) = \frac{n(E)}{n(S)} = \frac{2}{6} = \frac{1}{3}

 

 

Problem 1138

A single fair 80 sided die is rolled. Find the probability of getting a 5.

 

Solution:-

 

First, write the sample space S by listing all the possible outcomes for rolling a single die with 8 sides.

S = {1, 2, 3, 4, 5, 6, 7, 8}

The probability of an event E that is a sunset of a sample space S is P(E) = \frac{n(E)}{n(S)}, where n(E) is the number of outcomes in the event and n(S) is the number of outcomes in the sample space.

There are 8 outcomes in the sample space S = {1, 2, 3, 4, 5,6 , 7, 8}, so n(S) = 8.

Only 1 of these outcomes corresponds to getting a 5, so n(E) = 1.

Divide n(E) = 1 by n(S) = 8 to write the probability.

P(E) = \frac{n(E)}{n(S)} = \frac{1}{8}

Therefore, the probability of getting a 5 is \frac{1}{8}.