He would keep 3/8 for himself. Hope it helps! :)
Answer:
12
Step-by-step explanation:
Answer:
A) The best way to picture this problem is with a probability tree, with two steps.
The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.
The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.
If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.
If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.
B) So a person can have a red marble and a blue marble in two ways:
1) Picking the red first and the blue last
2) Picking the blue first and the red last
C) P(R&B) = 3/5 = 60%
Step-by-step explanation:
C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5
Answer:
Place a dot on 7 and do < underlined
Step-by-step explanation:
the probability is always the number of desired cases over the number of all possible cases.
in our situation we have 15 cards.
that is the total possible cases when a random card is chosen.
how many desired cases do we have ?
a number NOT a multiple of 5.
how many are there ?
it is easier to say how many numbers there are being a multiple of 5 : 5, 10, 15
so, 3 numbers out of the 15 are multiple of 5.
that means
15 - 3 = 12 numbers of the 15 are NOT multiples of 5.
so, the probability to draw a card that is not a multiple of 5 is
12/15 = 4/5 = 0.8
the information about event B and even numbers is irrelevant for the question.