A coin is tossed, and a standard number cube is rolled. What is the probability that the coin shows heads and the number cube shows an even number?
A. start fraction 1 over 6 end fraction
B. 1
C. one-fourth
D. one-half
There are 2 possible outcomes for the coin toss (heads or tails) and 3 possible outcomes for the number cube (even numbers are 2, 4, or 6). So there are 2 x 3 = 6 possible outcomes in total. Only one of those outcomes is heads and an even number: heads and 2. Therefore, the probability is 1/6.
Answer: A. start fraction 1 over 6 end fraction
A bag contains 3 blue marbles, 9 green marbles, and 11 yellow marbles. Twice you draw a marble and replace it. Find P(blue, then green).
A. start fraction 27 over 529 end fraction
B. start fraction 27 over 23 end fraction
C. start fraction 15 over 529 end fraction
D. Start Fraction 12 over 23 End Fraction
The probability of drawing a blue marble on the first draw is 3/23 (since there are 3 blue marbles out of 23 total). After replacing the first marble, the probabilities are reset, so the probability of drawing a green marble on the second draw is also 9/23.
To find the probability of both events happening (drawing blue then green), we multiply the probabilities:
P(blue, then green) = (3/23) x (9/23) = 27/529
So the answer is A. start fraction 27 over 529 end fraction.
The diagram shows the contents of a jar of marbles. You select two marbles at random. One marble is drawn and not replaced. Then a second marble is drawn. What is the probability of selecting a red marble and then another red marble?
An illustration of a jar of marbles is shown. Four of the marbles are labeled with an upper R, seven of the marbles are labeled with an upper B, and five of the marbles are labeled with an upper G. The key identifies upper R to represent red marbles, upper B to represent blue marbles, and upper G to represent green marbles.
A. start fraction 3 over 64 end fraction
B. one-fourth
C. Fraction is 1 over 20
D. start fraction 1 over 30 end fraction
There are 4 red marbles out of a total of 16 marbles. If one marble is drawn and not replaced, there will be a total of 15 marbles left in the jar, 3 of which are red. Therefore, the probability of drawing a red marble first is 4/16, or 1/4, and the probability of drawing another red marble second is 3/15.
To find the probability of both events happening (drawing a red marble and then another red marble), we multiply the probabilities:
P(red, then red) = (1/4) x (3/15) = 1/20
So the answer is C. Fraction is 1 over 20.
From a committee of 6 girls and 4 boys, a name is selected. Then another name is selected. What is the probability that both people drawn will be girls?
A. nine over twenty-five
B. two-fifteenths
C. three-fifths
D. one-third
There are 6 girls and 4 boys on the committee, so the probability of selecting a girl on the first draw is 6/10 and the probability of selecting another girl on the second draw (assuming the first selection is not replaced) is 5/9, since there will be one less person overall and one less girl.
To find the probability of both events happening (drawing two girls), we multiply the probabilities:
P(girl, then girl) = (6/10) x (5/9) = 1/3
So the answer is D. one-third.
Write the number of permutations in factorial form. Then simplify.
J K L M N O P
A. 7!; 720
B. 8!: 40,320
C. 7!; 49
D. 7!; 5,040
There are 7 letters in the given set, so the number of permutations is 7!.
7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5,040
So the answer is D. 7!; 5,040.
How many 2-digit numbers can be formed using only the digits 2, 3, 5, and 6, if the digits are not be repeated within a number?
A. 11
B. 12
C. 10
D. 2