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A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

One approach is to calculate the probability of flipping no heads, one head, two heads, etc., all the way to six heads, and adding those probabilities together, but that would be time consuming. Rather, calculate the probability of flipping seven heads. The complement to that would then be the sum of all other flip probabilities, which is what the problem calls for:

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

A coin is flipped four times. What is the probability of getting heads at least three times?

$\frac{5}{16}$

$\frac{1}{16}$

$\frac{1}{4}$

$\frac{11}{16}$

$\frac{3}{4}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

Per the question, we're looking for the probability of at least three heads; three head flips or four head flips would satisfy this:

$P(4,3)=\frac{4!}{(4-3)!3!}(\frac{1}{2})^{3}(\frac{1}{2})^{4-3}=4(\frac{1}{2})^{4}=\frac{1}{4}$

$P(4,4)=\frac{4!}{(4-4)!4!}(\frac{1}{2})^{4}(\frac{1}{2})^{4-4}=(\frac{1}{2})^{4}=\frac{1}{16}$

Thus the probability of three or more flips is:

$\frac{1}{4}+\frac{1}{16}=\frac{5}{16}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

A coin is flipped seven times. What is the probability of getting heads six or fewer times?

$\frac{127}{128}$

$\frac{255}{256}$

$\frac{1}{128}$

$\frac{1}{7}$

$\frac{6}{7}$

Explanation

Since this problem deals with a probability with two potential outcomes, it is a binomial distribution, and so the probability of an event is given as:

$P(n,k)=\frac{n!}{(n-k)!k!}p^k(1-p)^{n-k}$

Where is the number of events, is the number of "successes" (in this case, a "heads" outcome), and is the probability of success (in this case, fifty percent).

$P(7,7)=\frac{7!}{(7-7)!7!}(\frac{1}{2})^7(1-\frac{1}{2})^{7-7}=(\frac{1}{2})^7=\frac{1}{128}$

Therefore, the probability of six or fewer heads is:

$1-\frac{1}{128}=\frac{127}{128}$

Presented with a deck of fifty-two cards (no jokers), what is the probability of drawing either a face card or a spade?

$\frac{11}{26}$

$\frac{25}{52}$

$\frac{156}{2704}$

$\frac{3}{52}$

$\frac{9}{26}$

Explanation

A face card constitutes a Jack, Queen, or King, and there are twelve in a deck, so the probability of drawing a face card is $P(F)=\frac{12}{52}$ .

There are thirteen spades in the deck, so the probability of drawing a spade is $P(S)=\frac{13}{52}$ .

Keep in mind that there are also three cards that fit into both categories: the Jack, Queen, and King of Spades; the probability of drawing one is $P(F\bigcap S)=\frac{3}{52}$

Thus the probability of drawing a face card or a spade is:

$P(F)+P(S)-P(F\bigcap S)=\frac{12+13-3}{52}=\frac{11}{26}$

$\frac{5}{6}$ of the population in Town A is NOT Asian. In addition, $\frac{2}{3}$ of the Asian population in Town A is male. A random person from Town A is selected. What is the probability that the person selected is both Asian and female.

$\frac{1}{18}$

$\frac{10}{18}$

$\frac{1}{12}$

$\frac{11}{6}$

$\frac{1}{9}$

Explanation

Recall:

$\mbox{ ( the probability that an event occurs) = 1 - (the probability that the event doesn't occur) }$

Consequently, the probability that the person selected is Asian is:

$1 - \frac{5}{6} = \frac{6}{6} - \frac{5}{6} = \frac{1}{6}$

Similarly, the probability that a randomly selected Asian person is also female is:

$1 - \frac{2}{3} = \frac{1}{3}$

Finally, the probability of 2 unrelated events occuring is equal to the product of the individual probabilities of the 2 events. Therefore the probability of selecting an Asian female is

$\frac{1}{6} \times \frac{1}{3} = \frac{1}{18}$

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