1. Suppose a die with six sides (1,2,3,4,5,6) is tossed. (a.) What is the probab

Question

1. Suppose a die with six sides (1,2,3,4,5,6) is tossed.
(a.) What is the probability that the number that appears is a 2?
(b.) What is the probability that the number that appears is divisible by 3?
(c.) If the die is tossed twice, find the probability that the sum of the numbers appearing is at least 6.
2. Suppose a fair coin is tossed 4 times.
(a.) Write out all possible outcomes (i.e. HHTT, etc).
(b.) What is the probability of obtaining four tails?
(c.) What is the probability of obtaining two heads and two tails?
(d.) What is the probability of no heads?
(e.) What is the probability of obtaining at least 1 head?
3. A bag contains 4 blue marbles, 2 green marbles, 1 white marble and 5 red marbles.
(a.) Suppose a marble is selected from the bag and replaced. What is the probability of receiving a
blue marble on the first selection and a blue marble on the second selection?
(b.) Suppose a marble is selected from the bag and replaced. What is the probability of receiving a
blue marble on the first selection and not a blue marble on the second selection?
(c.) Suppose a marble is selected from the bag and replaced. What is the probability of receiving at
least one blue marble and a white marble (in either order)?
(d.) Suppose a marble is selected from the bag and not replaced. What is the probability of receiving
a blue marble on the first selection and a blue marble on the second selection?

Explanation / Answer

1. a)- probability that the number that appears is a 2= 1/6
b)- probability that the number that appears is divisible by 3= 2/6= 1/3
c)- for sum= 6 we can have following= (1,5) (2, 4), (3, 3) (5, 1) (4, 2)= 5 outcomes, and total outcomes= 36
so probability= 5/36

2. a)- possible outcomes= HHHH
HHHT
HHTH
HHTT
HTHH
HTTH
HTHT
HTTT
TTTT
TTTH
TTHT
TTHH
THTT
THTH
THHT
THHH

b)- probability of obtaining four tails = 1/36
c)- probability of obtaining two heads and two tails= 6/36 = 1/6
d)- probability of no heads= 1/36
e)- probability of obtaining at least 1 head= 1- probability of no heads= 1-1/36= 35/36

3. a) - bag contains 4 blue marbles, 2 green marbles, 1 white marble and 5 red marbles.
probability of receiving a blue marble on the first selection and a blue marble on the second selection
= 4/12 * 4/12 = 1/9

b)- probability of receiving a blue marble on the first selection and not a blue marble on the second selection
(4/12) (1- 4/12)= 1/3 * 2/3= 2/9

c)- probability of receiving at least one blue marble and a white marble= (1- 4/12) * (1/12) = 2/36 = 1/18

d) - marble is selected from the bag and not replaced. the probability of receiving
a blue marble on the first selection and a blue marble on the second selection= 4/12 * ( 3/11) + 4/12 * ( 4/11)
= 1/11 + 4/33= 7/33

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