Question

if a pair of dice is thrown find the probability that the sum is neither 7 or 11

Answer 1

$\huge{\mathcal{\underline{\green{SOLUTION}}}}$

A pair of dice is rolled once

So the total number of possible outcomes

$= {6}^{2} = 36$

Let A be the event that the sum is 7

Let B be the event that the sum is 11

So the event points for the event A

$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)$

So the total number of possible outcomes for the event A is 6

So

$P(A) = \frac{6}{36} = \frac{1}{6}$

Now the event points for the event B

$(5,6),(6,5)$

So

$P(B) = \frac{2}{36} = \frac{1}{18}$

Now A ∩ B is the of getting sum of both 7 & 11

So the event points for the event A ∩ B is 0

So P( A ∩ B)

$= 1 - ( P(A) + P( B) - P( A ∩ B))$

$= 1 - \frac{1}{6} - \frac{1}{18} + 0$

$= \frac{14}{18}$

$= \frac{7}{9}$

SO THE REQUIRED PROBABILITY

$= \: P( {A}^{c}∩ \: { B}^{c} )$

$= P {(A∪B)}^{c}$

$= 1 - P(A ∪B \: )$

Answer 2

$A pair of dice is rolled once So the total number of possible outcomes = {6}^{2} = 36=6​2​​=36 Let A be the event that the sum is 7 Let B be the event that the sum is 11 So the event points for the event A (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) So the total number of possible outcomes for the event A is 6 So P(A) = \frac{6}{36} = \frac{1}{6}P(A)=​36​​6​​=​6​​1​​ Now the event points for the event B (5,6),(6,5)(5,6),(6,5) So P(B) = \frac{2}{36} = \frac{1}{18}P(B)=​36​​2​​=​18​​1​​ Now A ∩ B is the of getting sum of both 7 & 11 So the event points for the event A ∩ B is 0 So P( A ∩ B) = 1 - \frac{1}{6} - \frac{1}{18} + 0=1−​6​​1​​−​18​​1​​+0 = \frac{14}{18}=​18​​14​​ = \frac{7}{9}=​9​​7​​ SO THE REQUIRED PROBABILITY hope \: it \: helps \: dude$