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Probability Problems


Probability Problems
  • Page 1
 1.  
Find the probability of getting a sum 7 from 3 throws of a dice?
a.
$\frac{1}{36}$
b.
$\frac{1}{16}$
c.
$\frac{1}{9}$
d.
$\frac{1}{6}$


Solution:

 
In 3 throws of a dice n(s) = 6$\times $6$\times $6 = 216
Let E = getting a sum = {(3,4),(5,2),(6,1),(4,3),(2,5),(1,6)}

P(E) = $\frac{n(E)}{n(s)}$
P(E) = $\frac{6}{216}$
P(E) = $\frac{1}{36}$

Answer :
$\frac{1}{36}$

 


Correct answer : (1)
 2.  
In a bag you have 11 red,4 blue and 9 orange balls. Suppose you picked up 1 ball randomly. Find the probability that it is neither red nor orange?
a.
$\frac{1}{12}$
b.
$\frac{1}{2}$
c.
$\frac{1}{6}$
d.
$\frac{1}{3}$


Solution:

 
Given: total number of balls = 11 + 4 + 9 = 24
n(E) = 4 [blue ball drawn]

P(E) = $\frac{n(E)}{n(s)}$
P(E) = $\frac{4}{24}$
P(E) = $\frac{1}{6}$

Answer:
Probability that it is neither red nor orange = $\frac{1}{6}$

 


Correct answer : (3)
 3.  
In a bag there are numbered 1 to 20 tickets then 1 ticket is drawn randomly. Find the probability that ticket drawn has a number which is a multiple of 5 and 7?
a.
$\frac{1}{3}$
b.
$\frac{3}{20}$
c.
$\frac{3}{5}$
d.
$\frac{3}{10}$


Solution:

 
Given :  numbers S = {1,2,3,4,.......,18,19,20}
Let E = event of getting a multiple of 5 and 7 = {5,7,10,14,15,20}  
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{6}{20}$
P(E) = $\frac{3}{10}$

Answer:
The probability that ticket drawn has a number which is a multiple of 5 and 7 = $\frac{3}{10}$ 

 


Correct answer : (4)
 4.  
Find the probability of getting a face card from an ordinary deck of 52 playing cards?
a.
$\frac{1}{13}$
b.
$\frac{3}{13}$
c.
$\frac{4}{13}$
d.
$\frac{2}{13}$


Solution:

Given : playing cards n(S) = 52
and n(E)= event of getting a face card = 12
P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{12}{52}$
P(E) = $\frac{3}{13}$

Answer:
The probability of getting a face card = $\frac{3}{13}$ 



Correct answer : (2)
 5.  
Find the probability of getting a non-face card or a 7 card  from an ordinary deck of 52 playing cards? 
a.
$\frac{10}{13}$
b.
$\frac{1}{13}$
c.
$\frac{3}{13}$
d.
$\frac{7}{13}$


Solution:

 
Given : playing cards n(S) = 52
and n(non-face)= event of getting a non-face card = 40
n(7)=event of getting a 7 card = 4
n(non-face and 7) = 4/52

P(E) = $\frac{n(E)}{n(S)}$
P(non-face) = $\frac{40}{52}$
P(7) = $\frac{4}{52}$
P(non-face and 7) = $\frac{4}{52}$
P(non-face or 7) = P(non-face)+P(7)-P(non-face and 7)
P(non-face or 7) = $\frac{40}{52}$+ $\frac{4}{52}$- $\frac{4}{52}$
P(non-face or 7) =  $\frac{40}{52}$
P(non-face or 7) =  $\frac{10}{13}$

Answer :
The probability of getting a non-face card or a 7 card = $\frac{10}{13}$ 

 


Correct answer : (1)
 6.  
Tom tossed 2 coins. Find the probability of getting two heads?
a.
$\frac{1}{3}$
b.
$\frac{1}{2}$
c.
$\frac{1}{4}$
d.
$\frac{1}{6}$


Solution:

As we know each coin has two possible outcomes Head and tail
Here we have 2 coins
So, n(S) = {(H,T)(T,H)(H,H)(T,T)}
Getting two heads n(E)={(H,H)}

P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{1}{4}$

Answer:
The probability of getting heads = $\frac{1}{4}$ 



Correct answer : (3)
 7.  
A family has 3 children, if the probability of giving birth to a boy is same as the giving birth to a girl. Find the probability getting at least two boys?
a.
$\frac{1}{3}$
b.
$\frac{1}{2}$
c.
$\frac{1}{8}$
d.
$\frac{1}{4}$


Solution:

Given : 3 children
Children in order n(S) = {(BBB),(BBG),(BGB),(BGG),(GGG),(GGB),(GBG)(GBB)} = 8
Event of getting 2 boys n(E) = {(BBB),(BBG),(BGB),(GBB)} = 4

P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{4)}{8}$
P(E) = $\frac{1}{2}$

Answer:
The probability of getting at least 2 boys = P(E) = $\frac{1}{2}$  



Correct answer : (2)
 8.  
Find the probability of getting two numbers whose product is odd. If two dice are thrown simultaneously.
a.
$\frac{2)}{7}$
b.
$\frac{1)}{10}$
c.
$\frac{1)}{8}$
d.
$\frac{2)}{9}$


Solution:

Two dice thrown simultaneously, n(S) = 6$\times$ 6 = 36
Event of getting two numbers product odd n(E) = {(1,3),(1,5),(3,1),(3,3),(3,5),(5,1),(5,3),(5,5)} = 8

P(E)=$\frac{n(E)}{n(S)}$
P(E)=$\frac{8}{36}$
P(E)=$\frac{2}{9}$

Answer:
The probability of getting two numbers whose product is odd = $\frac{2)}{9}$




Correct answer : (4)
 9.  
In a box, there are 7 marbles, 3 black and 4 white. What is the probability that a black marble drawn?
a.
$\frac{1}{7}$
b.
$\frac{2}{7}$
c.
$\frac{4}{7}$
d.
$\frac{3}{7}$


Solution:

Given marbles n(S) = 7
Black marble n(E) =3 

P(E) = $\frac{n(E)}{n(S)}$
P(E) = $\frac{3}{7}$

Answer:
The probability of getting a black marble = $\frac{3}{7}$



Correct answer : (4)
 10.  
In a room there are 10 girls and 15 boys. Find the probability of selecting at least 2 girls and 1 boy, if 3 children are selected randomly?
a.
$\frac{23}{90}$
b.
$\frac{27}{92}$
c.
$\frac{29}{95}$
d.
$\frac{24}{93}$


Solution:

Let S = sample space
E = event of selecting 2 girls and 1 boy 

n(S) = $^{25}C_{3}$
n(S) = 2300

n1(E) = $^{20}C_{2}$ $\Times$$^{15}C_{1}$
n(E) = 675

Then,
P(E) = $\frac{n(E)}{n(S)}$ 
P(E) = $\frac{675}{2300}$
P(E) = $\frac{27}{92}$

Answer:
The probability of selecting at least 2 girls and 1 boy = $\frac{27}{92}$



Correct answer : (2)

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