dale placed six cARDS IN A BAG.QONE CARD IS LABELD 1,TWO CARDS ARE LABELD2,AND RTHREE CARDS ARE LABELD 3.FIND THE PROBABILITY OOF PICKING TWO CARDS IN THE ORDER OF THE NUBERS LISTED TH EFIRST CARD IS RETURNED BEFORE THE SECOND IS PICKED ..... P,(1,2)
P(3,NOT1)
p(NOT2,2)
P(1,3)
P(2,3)
P(2,NOT2)
I will do one for you, I am pretty sure you can do the rest.
P(3,not1):
to get a 3, Prob. = 3/6 or 1/2
to get "not 1", there are 5 of "not 1), so the prob. of "not 1" = 5/6
since you are returning the cards, each event is independent of the next,
so
the P(3,not 1) = 1/2 * 5/6 = 5/12
To find the probability of picking two cards in a specific order, you can follow these steps:
1. Determine the total number of cards in the bag. In this case, Dale placed six cards in the bag.
2. Examine the cards and identify the numbers assigned to each card. In this scenario, there is one card labeled 1, two cards labeled 2, and three cards labeled 3.
3. Identify the specific order of cards you want to determine the probability for. For example, in the case of P(3, not 1), you want to find the probability of selecting a 3 as the first card and a card that is not 1 as the second card.
4. Calculate the probability for each step of the desired order.
In the case of P(3, not 1):
- The probability of selecting a 3 as the first card is 3/6, as there are three 3-labeled cards out of the total six cards.
- The probability of selecting a card that is not 1 as the second card, given that the first card was a 3, is 5/6. This is because there are five cards left that are not labeled 1 out of the remaining six cards.
5. Multiply the probabilities of each step together to obtain the overall probability. In the case of P(3, not 1), you would multiply 3/6 and 5/6:
P(3, not 1) = (3/6) * (5/6) = 15/36 = 5/12
Following these steps, you can calculate the probabilities for the other scenarios as well.