Independent and dependent events practice
1. A jar contains 5 red marbles and 3 green marbles. What is the probability of selecting a red marble and then selecting another red marble without replacement?
Solution: The probability of selecting a red marble from the jar on the first pick is 5/8. After one red marble has been selected and removed from the jar, there are now 4 red marbles and 3 green marbles left in the jar. Therefore, the probability of selecting another red marble on the second pick is 4/7. To find the probability of both events occurring, we multiply the probabilities together: 5/8 x 4/7 = 5/14
2. A bag contains 6 blue marbles, 4 red marbles, and 2 green marbles. What is the probability of selecting a blue marble and then selecting a red marble without replacement?
Solution: The probability of selecting a blue marble from the bag on the first pick is 6/12, which can be simplified to 1/2. After one blue marble has been selected and removed from the bag, there are now 4 red marbles and 11 total marbles left in the bag. Therefore, the probability of selecting a red marble on the second pick is 4/11. To find the probability of both events occurring, we multiply the probabilities together: 1/2 x 4/11 = 2/11
3. A coin is flipped twice. What is the probability of getting heads on the first flip and tails on the second flip?
Solution: The probability of getting heads on the first flip is 1/2, and the probability of getting tails on the second flip is also 1/2. To calculate the probability of both events occurring, we multiply the probabilities together: 1/2 x 1/2 = 1/4
4. A deck of cards contains 52 cards, including 12 face cards (4 kings, 4 queens, and 4 jacks). If a card is randomly selected from the deck and then placed back in the deck, and then another card is randomly selected, what is the probability of selecting a king on the first pick and a queen on the second pick?
Solution: The probability of selecting a king on the first pick is 4/52, which can be simplified to 1/13. After the first card has been selected and replaced in the deck, there are still 4 kings and now 51 total cards in the deck. Therefore, the probability of selecting a queen on the second pick is 4/51. To find the probability of both events occurring, we multiply the probabilities together: 1/13 x 4/51 = 4/663
Sure! Here are some practice problems on independent and dependent events:
Problem 1: A bag contains 5 red marbles and 3 green marbles. If you randomly select 2 marbles without replacement, what is the probability that both marbles are red?
Solution:
To solve this problem, we need to determine if the events are independent or dependent. Since we are selecting marbles without replacement, the events are dependent.
Step 1: Find the probability of selecting a red marble on the first draw.
P(Red on the first draw) = Number of red marbles / Total number of marbles
= 5 / 8
Step 2: Find the probability of selecting a red marble on the second draw, given that the first marble was red.
P(Red on the second draw | first marble was red) = Number of remaining red marbles / Remaining marbles
= 4 / 7
Step 3: Multiply the probabilities from both steps to find the probability that both marbles are red.
P(Both marbles are red) = P(Red on the first draw) * P(Red on the second draw | first marble was red)
= (5 / 8) * (4 / 7)
= 20 / 56
So, the probability that both marbles are red is 20/56 or 5/14.
Problem 2: A deck of cards contains 52 cards, including 4 aces. If two cards are randomly drawn without replacement, what is the probability that the first card is an ace and the second card is also an ace?
Solution:
To solve this problem, we need to determine if the events are independent or dependent. Since we are selecting cards without replacement, the events are dependent.
Step 1: Find the probability of drawing an ace on the first draw.
P(Ace on the first draw) = Number of aces / Total number of cards
= 4 / 52
= 1/13
Step 2: Find the probability of drawing an ace on the second draw, given that the first card was an ace.
P(Ace on the second draw | first card was an ace) = Number of remaining aces / Remaining cards
= 3 / 51
= 1/17
Step 3: Multiply the probabilities from both steps to find the probability that both cards are aces.
P(Both cards are aces) = P(Ace on the first draw) * P(Ace on the second draw | first card was an ace)
= (1/13) * (1/17)
= 1 / 221
So, the probability that the first card is an ace and the second card is also an ace is 1/221.