To find the probability that both tickets are for games against the Snakes, we need to consider the total number of possible outcomes (2 games out of 59) and the number of favorable outcomes (2 games out of 4 against the Snakes).
Total possible outcomes = 59C2 (combinations of 2 out of 59) = 59! / (2!(59-2)!) = 1711
Favorable outcomes = 4C2 (combinations of 2 out of 4) = 4! / (2!(4-2)!) = 6
So, the probability is: Favorable outcomes / Total possible outcomes = 6 / 1711 = 6/1711
Therefore, the probability that both tickets are for games against the Snakes is 6/1711.
Your employer is offering free tickets for two baseball games. They play
59
59 home games and
4
4 of them are against the Snakes. The tickets are randomly given out. What is the probability that both are games are against the Snakes? Write your answer as a fraction. You do not need to simplify. Click Here for Help Video
5 answers
A jar has
16
16 marbles and
8
8 of them are green. You pull 3 marbles without replacement. What is the probability that all are green? Write your answer as a fraction. You do not need to simplify.
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16
16 marbles and
8
8 of them are green. You pull 3 marbles without replacement. What is the probability that all are green? Write your answer as a fraction. You do not need to simplify.
Click Here for Help Video
To find the probability that all three marbles are green, we need to consider the total number of possible outcomes (pulling 3 marbles out of 16) and the number of favorable outcomes (pulling all 3 green marbles out of the 8 green marbles).
Total possible outcomes = 16C3 (combinations of 3 out of 16) = 16! / (3!(16-3)!) = 560
Favorable outcomes = 8C3 (combinations of 3 out of 8) = 8! / (3!(8-3)!) = 56
So, the probability is: Favorable outcomes / Total possible outcomes = 56 / 560 = 56/560
Therefore, the probability that all three marbles pulled are green is 56/560.
Total possible outcomes = 16C3 (combinations of 3 out of 16) = 16! / (3!(16-3)!) = 560
Favorable outcomes = 8C3 (combinations of 3 out of 8) = 8! / (3!(8-3)!) = 56
So, the probability is: Favorable outcomes / Total possible outcomes = 56 / 560 = 56/560
Therefore, the probability that all three marbles pulled are green is 56/560.
A jar has
19
19 marbles and
7
7 of them are green. You pull 2 marbles without replacement. What is the probability that both are green? Write your answer as a fraction. You do not need to simplify. Click Here for Help Video
19
19 marbles and
7
7 of them are green. You pull 2 marbles without replacement. What is the probability that both are green? Write your answer as a fraction. You do not need to simplify. Click Here for Help Video
To find the probability that both marbles pulled are green, we need to consider the total number of possible outcomes (pulling 2 marbles out of 19) and the number of favorable outcomes (pulling both 2 green marbles out of the 7 green marbles).
Total possible outcomes = 19C2 (combinations of 2 out of 19) = 19! / (2!(19-2)!) = 171
Favorable outcomes = 7C2 (combinations of 2 out of 7) = 7! / (2!(7-2)!) = 21
So, the probability is: Favorable outcomes / Total possible outcomes = 21 / 171 = 21/171
Therefore, the probability that both marbles pulled are green is 21/171.
Total possible outcomes = 19C2 (combinations of 2 out of 19) = 19! / (2!(19-2)!) = 171
Favorable outcomes = 7C2 (combinations of 2 out of 7) = 7! / (2!(7-2)!) = 21
So, the probability is: Favorable outcomes / Total possible outcomes = 21 / 171 = 21/171
Therefore, the probability that both marbles pulled are green is 21/171.
1 answer