1. A spinner with 6 equal sections numbered 1 – 6 is spun once. What is the probability of getting each of the following:
a. an even number greater than 3 _____________________________
b. an odd number less than 5 _______________________________
c. a prime number less than 3 ________________________________
*Hint: A prime number is a number whose only factors are 1 and itself.
d. a 3 or a 6_______________________________________...
2. A single candy is randomly picked from a bag containing 4 red candies, 5 green candies, and 1 blue candy. Find the probability that the candy picked is:
a. blue ________________________________________...
b. either red or green ______________________________________
c. not green ________________________________________...
d. neither green nor blue ___________________________________
e. white ________________________________________...
3. One card is drawn from a standard deck of 52 playing cards. Find the probability that it is:
a. a 3 or a 7 ________________________________________...
b. a 9 or a heart___________________________________...
c. either a heart or a diamond_________________________________...
d. neither an ace nor a queen ___________________________________
4. If you meet a family with three children, what is the probability that there are exactly two girls? *Hint: Draw a tree diagram
5. If you toss a coin, what is the probability of getting each of the following outcomes?
a. all heads___________________________________...
b. no heads___________________________________...
c. exactly one head ________________________________________...
d. exactly 2 tails ________________________________________...
6. Corrine has 3 quarters, 7 dimes, 6 nickels, and 4 pennies in her pocket. If one coin falls out, what is the probability that it is one of the following?
a. either a nickel or a dime____________________________________...
b. worth exactly 10 cents___________________________________...
c. worth more than 10 cents___________________________________...
d. worth more than 25 cents___________________________________...
7. Carol reached into her closet and without looking, pulled a book off the shelf. There were 4 books, science, math, Spanish and English. Without replacing the first, she picked a second book.
a. What is the probability that she chose the English book followed by the Spanish book?
______________________________________...
b. What is the probability that she chose two different books?
______________________________________...
8. A bag contains 2 red cubes, 3 blue cubes, and 5 green cubes. If a cube is removed, examined, and replaced in the bag, what is the probability that, after two draws:
a. both cubes are red ________________________________________...
b. the first is red and the second is blue______________________________
c. both are green ________________________________________...
d. neither is blue ________________________________________...
Help with probability..can u tutor me maybe?
Question 1:
(a) E = {even number greater than 3} = {4,6}
P(E) = 1/6 + 1/6 = 1/3
(b) E = {odd number less than 5} = {1,3}
P(E) = 1/6 + 1/6 = 1/3
(c) E = {prime number less than 3} = {2}
*1 is not a prime number, and this doesn't include 3
P(E) = 1/6
(d) E = {3 or 6} = {3, 6}
P(E) = 1/6 + 1/6 = 1/3
Question 2:
Given,
4R (four red candies), 5G (five green candies), 1B (one blue candy); and ONE candy is drawn
(a) It's a blue candy, P(B) = n(B) / n(C)
{Probability of a blue candy equals the number of blue candies over total number of candies}
P(B) = 1 /10
(b) Either red or green, P(R or G)
= P(R) + P(G)
= 4/10 + 5/10
= 9/10
(c) Not green, so it must be P(R or B)
= P(R) + P(B)
= 4/10 + 1/10
= 5/10 = 1/2
(d) Not green or blue, so it must be P(R)
= 4/10 = 2/5
(e) White, there is no white in, so
P(W) = 0
Question 3:
Given, a standard deck of cards has 52 cards,
n(S) = 52
(a) a 3 or a 7, P (3 or 7)
= P(3) + P(7)
Now, each number only appears four times in a standard deck of cards. For example, 3 appears only as 3 of diamonds, 3 of hearts, 3 of clubs and 3 of spades.
= 4/52 + 4/52
= 8/52
(b) a 9 or a heart, P(9 or heart)
= P(9) + P(heart)
Now, the patterns of the cards, appear 13 times each. There's only 4 patterns: hearts, diamonds, clubs, spades. 52 cards, 4 patters; that means 13 cards each. You can count them: Ace (1) to (10) then your Jack, Queen, King. 13. So,
= 4/52 + 13/52
= 17/52
(c) a heart or a diamond, P(heart or diamond)
= P(heart) + P(diamond)
= 13/52 + 13/52
= 26/52 = 1/2
(d) not an Ace or a Queen, P(notAce or notQueen)
= P(notAce) + P(notQueen)
= 1 - P(Ace) - P(Queen)
= 1 - 4/52 - 4/52
= 1 - 8/52
= 44/52 = 11/13
Question 4:
Given, a family with three children. So the combinations are:
BBB
GGG
BGG
GBB
In the case of BBB and GGG,
There can be NONE, or TWO who look identical or ALL THREE looking identical. So, each combination has three probabilities.
In the case of BGG or GBB however,
There can be NONE, or all THREE who look identical. But in the case of the TWO who look identical, it can be a match of each sex or it can be both of the same sex. One boy and one girl or two girls. So, each combination has four probabilities.
We will only use the combination with two girls or more:
BGG and GGG
P(two identical girls)
= P(two identical in GGG) + P(two identical girls in BGG)
= [ (1/4) x (1/3) ] + [ (1/4) x (1/4) ]
= 1/12 + 1/16
= 7/48
Question 5, I don't quite know how to answer you when you did not supply me with the number of times the coin was tossed.
Question 6:
Given, 3Q (three quarters), 7D (seven dimes), 6N (six nickels) and 4P (four pennies); and ONE coin falls out.
(a) either a nickel or a dime, P(N or D)
= P(N) + P(D)
= 6/20 + 7/20
= 13/20
(b) exactly 10 cents, P(D)
= 7/20
(c) worth %26gt;10 cents, P(Q)
= 3/20
(d) worth %26gt;25 cents, since quarters are the biggest value, and there's nothing bigger than that in the pocket,
P(%26gt;25 cents) = 0
Question 7:
I'm assuming it's 1 Science book, 1 Math book, 1 Spanish book and 1 English book.
(a) Without replacement, so it's P(E and S)
= P(E) x P(S)
= 1/4 x 1/4
= 1/16
(b) Since there are four books only, and all different-
P(two different books) = 1
There is only one outcome- all of them will be different books.
Question 8:
Given, 2R (two reds), 3B (three blues) and 5G (five greens); and ONE cube is taken and replaced, then another ONE cube is taken.
(a) both are red, P(R and R)
since it's replaced, they're independant events.,
= P(R) x P(R)
= 2/10 x 2/10
= 4/100
= 1/25
(b) first is red, second is blue, P(R and B)
= P(R) xP(B)
= 2/10 x 3/10
= 6/100
= 3/50
(c) both are green, P(G and G)
= P(G) x P(G)
= 5/10 x 5/10
= 25/100
= 1/4
(d) neither is blue, P(notB and notB)
P(notB)
= P(G) + P(R)
= 5/10 + 2/10
= 7/10
P(not B) x P(not B)
= 7/10 x 7/10
= 49/100
I haven't done Math in a long time so I may be a bit rusty.
Reply:all that is actually REALLY easy. in the yahoo page, write PERMUTATIONS. thats the symbol (!) theres a lot of questions like this on the SAT. just trust me, wrirte permutttions
gladiolus
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