SAT Math - How to find the probability of an outcome

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:

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

Mike has a bag of marbles, 4 white, 8 blue, and 6 red. He pulls out one marble from the bag and it is red. What is the probability that the second marble he pulls out of the bag is white?

4/18

3/18

4/17

1/6

Explanation

There are 18 marbles in total. One of them is removed so now there are 17 marbles. This is our denominator. All of the original white marbles are still in the bag so there is a 4 out of 17 or 4/17 chance that the next marble taken out of the bag will be white.

$\frac{5}{6}$ 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

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

If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?

3/10

9/20

6/10

9/10

11/20

Explanation

If x is chosen at random from the set (4, 6, 7, 9, 11) and y is chosen at random from the set (12, 13, 15, 17) then what is the probability that xy is odd?

Here we have 5 possible choices for x and 4 possible choices for y, giving us 5 * 4 = 20 possible outcomes.

We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where x and y are odd. We have 3 possibilities of odd numbers for x, and 3 possibilities of odd numbers for y, so we will have 9 outcomes of our total 20 outcomes where xy is odd, giving us a probability of 9/20.

In a bag, there are 6 black and 8 yellow marbles. A yellow marble is chosen at random and is not replaced. What is the probability of choosing a second yellow marble?

$\dpi{100} \small \frac{7}{13}$

$\dpi{100} \small \frac{4}{7}$

$\dpi{100} \small \frac{6}{13}$

$\dpi{100} \small \frac{6}{7}$

Explanation

Because a yellow marble was selected and not replaced, there is a total of 13 marbles in the bag and of which, 7 are yellow; therefore, the probability of selecting a yellow marble is $\dpi{100} \small \frac{7}{13}$

10 cards, each with a distinct number from 1-10, are placed face-down on a table. If Matt chooses a card at random, what is the probability his card will display a prime number?

$\small \frac{2}{5}$

$\small \frac{1}{10}$

$\small \frac{1}{5}$

$\small \frac{3}{10}$

$\small \frac{5}{10}$

Explanation

Total outcomes: 10

Desired outcomes (prime numbers) = 4 (2, 3, 5, 7)

$\small \frac{4}{10}:=:\frac{2}{5}$

If 2 six-sided dice, each with sides numbered 1-6, are rolled, what is the probability that the sum of the numbers on the face-up sides is equal to 7?

$\small \frac{1}{6}$

$\small \frac{1}{3}$

$\small \frac{1}{12}$

$\small \frac{5}{36}$

$\small \frac{1}{4}$

Explanation

Probability = desired outcomes divided by total possible outcomes.

Desired outcomes = 6 (1,6; 2,5; 3,4; 4,3; 5,2; 6;1)

Total outcomes - 36 $\left ( 6\times 6 \right )$

$\small \frac{6}{36}:=:\frac{1}{6}$

If given two dice, what is the probability that the sum of the two numbers rolled will equal 9?

1/6

1/9

1/18

1/24

1/36

Explanation

There are 36 possible outcomes of the additive dice roll. The way to roll a sum of 9 is 6 (and vice versa) and 3 or 5 and 4 (and vice versa). This is possible 4 of the 36 times, giving a probability the sum of the two dice rolled of 4/36 or 1/9.

How to find the probability of an outcome

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