six cards are numbered 1 1 6 7 11 and 12 cards are chosen without replacement until a card number 1 is chosen. find the probability that this happens before the third card is chosen
Well, imagine these cards are on a blind date. The odds of a "1" card being chosen on the first try are 2 out of 6, which simplifies to 1 out of 3, or 33.33333...%.
If a "1" card is not drawn on the first try, the odds of a "1" card being chosen on the second try go down. Now, there are 5 cards remaining, but only 1 of them is a "1" card. So the odds drop to 1 out of 5, or 20%.
Finally, if a "1" card is not drawn on the second try, there are only 4 cards left, but still only 1 of them is a "1" card. So the odds decrease to 1 out of 4, or 25%.
To find the probability that a "1" card is chosen before the third card is drawn, we need to multiply the probabilities of each event happening in sequence:
1/3 × 4/5 × 1/4 = 4/60 = 1/15
So the probability is 1 out of 15, or approximately 6.666666...%.
To find the probability that a card numbered 1 is chosen before the third card, we need to consider the possible arrangements.
Let's analyze the different scenarios step by step:
1. The card "1" is chosen as the first card:
The probability of this happening is 2/6, as there are two cards numbered 1 out of a total of six cards.
2. The card "1" is chosen as the second card:
For this scenario to happen, the first card chosen must not be numbered 1. The probability of this happening is 4/6.
Then, the second card chosen must be numbered 1. The probability of this happening is 2/5.
So the probability of this scenario is (4/6) * (2/5) = 8/30.
3. The card "1" is chosen as the third card:
For this to occur, the first two cards chosen must not be numbered 1. The probability of this happening is (4/6) * (3/5).
Then, the third card chosen must be numbered 1. The probability of this happening is 2/4.
So the probability of this scenario is (4/6) * (3/5) * (2/4) = 24/120.
Since we are interested in finding the probability of the event happening before the third card is chosen, we need to add up the probabilities from scenarios 1 and 2:
2/6 + 8/30 = 10/15 = 2/3
Therefore, the probability of choosing a card numbered 1 before the third card is chosen is 2/3.
To find the probability that a card numbered 1 is chosen before the third card is chosen, we need to consider the possible outcomes and count the favorable outcomes.
There are two possible outcomes:
1) A card numbered 1 is chosen before the third card is drawn.
2) A card numbered 1 is not chosen before the third card is drawn.
Let's calculate the favorable outcomes for each of these cases:
Case 1: A card numbered 1 is chosen before the third card is drawn.
There are only three cards numbered 1 among the six cards selected. So, we can find the favorable outcomes by considering the positions of these three cards. The possible arrangements are:
1 1 X X X X
X 1 1 X X X
X X 1 1 X X
X X X 1 1 X
X X X X 1 1
X X 1 X 1 X
X X 1 X X 1
X 1 1 X 1 X
X 1 1 X X 1
X 1 X 1 X X
X 1 X X 1 X
X X 1 1 X X
There are 11 favorable outcomes for Case 1.
Case 2: A card numbered 1 is not chosen before the third card is drawn.
To calculate the favorable outcomes for this case, we need to find the number of ways to choose 3 cards from the remaining 9 cards (excluding the three cards numbered 1). This can be calculated using combinations.
The number of ways to choose 3 cards from 9 is given by the equation:
9C3 = (9!)/((3!)(9-3)!) = 84
So, there are 84 favorable outcomes for Case 2.
Therefore, the total number of favorable outcomes is 11 + 84 = 95.
To find the probability, we need to divide the number of favorable outcomes by the total number of possible outcomes, which is the number of ways to choose 12 cards from 6:
6C12 = (6!)/((12!)(6-12)!) = 924
The probability is:
P = (Number of favorable outcomes) / (Number of possible outcomes) = 95/924 ≈ 0.103
Therefore, the probability that a card numbered 1 is chosen before the third card is chosen is approximately 0.103.
"Before the third card is chosen" means a 1 is chosen as either the first card or second card.
Probability of 1 being chosen as first card = 2/6.
Assuming that the first card was not a 1, the probability of 1 being chosen as a second card = 2/5.
Either-or probabilities are found by adding the individual probabilities.