Question

Each of the faces of a fair six-sided number cube is numbered with one of the numbers 1 through 6, with a different number appearing on each face. Two such number cubes will be tossed, and the sum of the numbers appearing on the faces that land up will be recorded. What is the probability that the sum will be 4, given that the sum is less than or equal to 6 ?

Answer 1

Answer:

0.2

Step-by-step explanation:

If two number cubes as tossed then the total possible out comes are

Total = {(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)}

Number of elements in sample space are

n(S)=36

Let A and B represent the following events.

A = The sum will be 4.

A = {(1,3),(2,2),(3,1)}

n(A) = 3

B = The sum is less than or equal to 6.

B = {(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

n(B)=15

Intersection of both events is

A∩B = {(1,3),(2,2),(3,1)}

n(A∩B) = 3

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{36}$

$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{15}{36}$

$P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{3}{36}$

We need to find the probability that the sum will be 4, given that the sum is less than or equal to 6.

$P(\frac{A}{B})=\dfrac{P(A\cap B)}{P(B)}$

$P(\frac{A}{B})=\dfrac{\frac{3}{36}}{\frac{15}{36}}$

$P(\frac{A}{B})=\dfrac{3}{15}$

$P(\frac{A}{B})=\dfrac{1}{5}$

$P(\frac{A}{B})=0.2$

Therefore, the required probability is 0.2.

Answer 2

Answer:

The probability that the sum will be 4 = 0.2

Step-by-step explanation:

When two such number cubes will be tossed than total possible outcomes are = {(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Sample space has total number of elements,

n(S) = 36

Let us assume that C and D are that events

Sum will be 4 = C

C = {(1,3)(2,2)(3,1)}

n(C) = 3

Sum will be less than or equal to 6 = D

D = {(1,1),(1,2),(1,3),(1,4),(1,5),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(4,1),(4,2),(5,1)}

n(D) = 15

Both events has intersection,

$\rm C \cap D$ = {(1,3)(2,2)(3,1)}

$\rm n (C \cap D)$ = 3

$\rm P(C) = \dfrac {n(C)}{n(S)} = \dfrac{3}{36}$

$\rm P(D) = \dfrac {n(D)}{n(S)} = \dfrac{15}{36}$

$\rm P(C \cap D) = \dfrac {n(C \cap D)}{n(S)} = \dfrac{3}{36}$

The Probability that the sum will be 4 and given that the sum will less rthan equal to 6,

$\rm P(\frac {C}{D}) = \dfrac{P(C \cap D)}{P(D)}$

$\rm P(\frac {C}{D}) = \dfrac{\frac{3}{36}}{\frac{15}{36}} = \dfrac {1}{5}$

$\rm P(\frac{C}{D}) = 0.2$

Therefore, the probability that the sum will be 4 = 0.2

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