PROBABILITY QUESTIONS WITH SOLUTIONS

1.The probability that a student will get an A in statistics is .20, the probability that she will get an A in French is .25, and the probability that she will get an A in both is .05. What is the probability that she will get an A in neither subject?

 solution: 

To begin, we need to recognize that P(A in neither subject) = 1 - P(A in Statistics or A in French)
So, we need to find P(A in Statistics or A in French)

We'll use the formula P(X or Y) = P(X) + P(Y) - P(X and Y)
So, P(A in Statistics or A in French) = P(A in Statistics) + P(A in French) - P(A in Statistics and A in French)
= 0.2 + 0.25 - 0.05
So, P(A in neither subject) = 1 - 0.4
= 0.6

Or

Probability of getting A in Statistics --> 0.20
Probability of getting A in French --> 0.25
Probability of getting A in both --> 0.05

Total Probability = Probability of getting A only in Statistics + Probability of getting A only in French + Probability of getting A in both + Probability of getting A in neither = 1

i.e. 1 = (0.20 - 0.05) + (0.25 - 0.05) + 0.05 + x
--> x = 1 - 0.15 - 0.20 - 0.05 = 0.6

2.Eight world leaders are lined up randomly to be photographed at an international summit. If the leaders of Brazil and Hungary are among the eight leaders, what is the probability that they will be standing next to each other in the photograph?

A 1/64    B 3/32     C 1/8     D 1/4     E 1/2

If one of the delegates ends up in one of the six middle spots (which has probability 3/4), the probability the other will be next to them is 2/7, but if one is placed on the end (which has probability 1/4), the probability of them being next to each other is only 1/7.

So the total probability is (1/4)(1/7)+(3/4)(2/7) = 7/28 = 1/4

==> Answer is D.

Or

suppose leaders are L₁, L₂, L₃, L₄, L₅, L₆, L₇ and L₈

all possible formation = ⁸P₈ = 8!

from 8 positions, we can choose 2 side-by-side positions in 7 ways -
i.e. L₁-L₂, L₂-L₃, L₃-L₄ ...

these 2 positions will be used by 2 ways - brazil-hungary and hungary-brazil

we have 6 positions remained, now 6 people can be placed in to 6 places = ⁶P₆ = 6!

probability = (2 X 7 X 6!)/8! = (2 X 7!)/(8 X 7!) = ¼

or

Take those two individual as ONE. SO there are 7 people who can be arranged in 7! ways. And 8 people can be arranged in 8! ways. Again those two can arranged themselves in 2 ways. So the Probability is 2 X (7!/8!) = 2 / 8 = ¼

1.There are 9 beads in a bag. 3 beads are red, 3 beads are blue, and 3 beads are black. If two beads are chosen at random, what is the probability that they are both blue?

A. 1/81   B. 1/12   C. 2/9    D. 1/3    E. ¼

Sol B
Probability of getting one blue bead = 3/9 = 1/3
Porbability of getting second blue bead = 2/8 = 1/4
Probability of getting blue bead in both picks = 1/3*1/4 = 1/12

2 A letter is randomly selected from the word Mississippi. What is the probability that the letter will be an s?

A. 1/11   B. 3/10    C. 4/11    D. 1/4    E. 1/3

Sol: 
Probability of choosing s = 4/11

3. A certain deck of cards contains 2 blue cards, 2 red cards, 2 yellow cards, and 2 green cards. If two cards are randomly drawn from the deck, what is the probability that they will both are not blue?

A. 15/28    B. 1/4   C. 9/16   D. 1/32   E. 1/16

Sol:
 A
Probability for first card to be not blue - 6/8 = 3/4
Probability for second card to be not blue - 5/7
Probablility for both cards to be not blue - 3/4*5/7 = 15/28

4. A fair coin is tossed, and a fair six-sided die is rolled. What is the probability that the coin come up heads and the die will come up 1 or 2?

A. 1/2    B. 1/4   C. 1/6     D. 1/12     E. 1/3

Sol:  C
Probability for heads - 1/2
Probability for 1 or 2 - 1/3
Probability for heads and 1 or 2 - 1/2*1/3 = 1/6

5. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is theprobability that at least one marble is blue?

A. 21/50     B. 3/13   C. 47/50   D. 14/15   E. 1/5

Sol: . D
Atleast on blue marble can be selected in three ways, The GRE Big Book, RB and BR
Probability of getting The GRE Big Book - 7/10*6/9 = 7/15
Probability of getting RB - 3/10*7/9 = 7/30
Probability of getting BR - 7/10*3/9 = 7/30
Probability of getting atlease on blue - 7/15 + 7/30 + 7/30 = 14/15

6. A fair, six-sided die is rolled. What is the probability that the number will be odd?

A. 1/4     B. 1/2   C. 1/3   D. 1/6    E. 1/5

Sol:
B
There are three odd numbers 1,3 and 5
Probability of getting odd number - 3/6 = 1/2

7. A letter is randomly select from the word studious. What is the probability that the letter be a U?

A. 1/8    B. 1/4    C. 1/3    D. 1/2    E. 3/8

Sol:
 B
Probability of choosing u - 2/8 = 1/4

8. A bag contains 2 red beads, 2 blue beads, and 2 green beads. Sara randomly draws a bead from the bag, and then Victor randomly draws a bead from the bag. What is the probability that Sara will draw a red marble and Victor will draw a blue marble?

A. 2/13    B. 1/5      C. 1/3     D. 1/10      E. 2/15

Sol:
E
Probability that Sara draws a red marble - 2/6 = 1/3
Probability that victor draws a blue marble - 2/5
Probability that Sara draws a blue marble and Victor draws a red marble - 1/3 x 2/5 = 2/15

9. If two fair, six-sided dice are rolled, what is the probability that the sum of the numbers will be 5?

A. 1/6     B. 1/4     C. 1/36    D. 1/18    E. 1/99.
sol:There are 36 pairs of numbers when two dice are rolled. There four combinations that gives the sum of 5 - (1,4), (2,3), (3,2) and
(4,1)
Probability of the sum to be 5 - 4/36 = 1/9

10. If four fair coins are tossed, what is the probability of all four coming up heads?

A. 1/4       B. 1/6       C. 1/8     D. 1/16     E. 1/32

Sol: D
Probability of getting all four heads - 1/2 x 1/2 x 1/2 x 1/2 = 1/16

11. The probability that a certain event will occur is 5/9. What is the probability that the event will NOT occur?

A. 5/9   B. 4/9     C. 2/9    D. 1/4     E. 1/2
sol: B
Probability of the event not to occur = 1 - 5/9 = 4/9

12. A certain bag contains red, blue, yellow, and green marbles. If a marble is randomly drawn from the bag, the probabilityof drawing a blue marble is .2, the probability of drawing a red marble is .3, and the probability of drawing a yellow marble is .1. What is the probability of drawing a green marble?

A. .5      B. .6      C. .2     D. .4       E. .3

Sol: D
Probability of drawing green marble = 1 - (0.2 + 0.3 + 0.1) = 0.4

13. A bag contains 3 red marbles, 3 blue marbles, and 3 green marbles. If a marble is randomly drawn from the bag and a fair, six-sided die is tossed, what is the probability of obtaining a red marble and a 6?

A. 1/15    B. 1/6     C. 1/3     D. 1/4      E. 1/18

Sol:  E
Probability getting red marble = 3/9 = 1/3
Probability of getting 6 = 1/6
Probability of getting red marble and 6 = 1/3 x 1/6 = 1/18

14. A fair, six-sided die is rolled. What is the probability of obtaining a 3 or an odd number?

A. 1/6

B. 1/5

C. 1/4

D. 2/3

E. ½

Sol:  E
As 3 itself is an odd number, probability of getting odd number = 3/6 = 1/2

15. At a certain business school, 400 students are members of the sailing club, the wine club, or both. If 200 students are members of the wine club and 50 students are members of both clubs, what is the probability that a student chosen at random is a member of the sailing club?

A. 1/2      B. 5/8       C. 1/4      D. 3/8       E. 3/5

Sol: B
Members in sailing club = 250
Probability of choosing member from sailing club = 250/400 = 5/8

16. A bag contains six marbles: two red, two blue, and two green. If two marbles are drawn at random, what is theprobability that they are the same color?

A. 1/3       B. 1/2      C. 1/8     D. 1/4       E. 1/5

Sol:  E
First marble can be any color.
Probability of choosing second marble of the same color as the first one = 1/5

17. There are five students in a study group: two finance majors and three accounting majors. If two students are chosen at random, what is the probability that they are both accounting students?

A. 3/10     B. 2/5        C. 1/5       D. 3/5        E. 4/5

sol:Probability of first student to be accounting student - 3/5
Probability of second student to be accounting student - 2/4 = 1/2
Probability that both students to be accounting students - 3/5 x 1/2 = 3/10

18. Seven beads are in a bag: three blue, two red, and two green. If three beads are randomly drawn from the bag, what is the probability that they are not all blue?

A. 5/7          B. 23/24      C. 6/7      D. 34/35      E. 8/13

sol; D
Probability of selecting all blue marbles in 3 picks - 3/7 x 2/6 x 1/5 = 1/35
Probability that all three of them are not blue = 1 - 1/35 = 34/35

19. A bag has six red marbles and six blue marbles. If two marbles are drawn randomly from the bag, what is the probabilitythat they will both be red?

A. 1/2        B. 11/12     C. 5/12       D. 5/22      E. 1/3

sol:

 D
Probability that both are red marbles = 6/12 x 5/11 = 5/22