# Find the conditional probability in a single roll of two fair dice, that a)The sum is less than 6, given that the sum is even b)The sum is 10, given that the roll is doubles c)The sum is even, given that the sum is less than 6 d)The roll is doubles,given that the sum is 10 e)The sum is greater than 7, given that neither die is a six f)The sum is odd, given that at least one die is a six g)Neither die is a six, given that the sum is greater than 7 h)At least one die is a six, given that the sum is odd

2
by sweetysiri92
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2015-03-23T20:01:41+05:30

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P(B | A) = probability of occurrance of event B , given that event A has occurred
= P(A П B) / P(A)
A Π B  means the intersection of events A and B, occurring together.
We also know that  if  P (B | A ) = P(B)   ie.,  P (A Π B) = P(A) * P(B)  the two events A and B are independent of each other.

Dice D1 and D2 are rolled once.  Let the numbers displayed on the dice be d1 and d2 respectively.  The dice D1 and D2 are independent.
==================
a)
Sum S =  d1 + d2  = even, for  d1 = d2 = odd   OR    d1 = d2 = even
P(d1 = even) = 3/6 =1/2      P(d2=even) = 1/2
P(d1 = odd) = 1/2              P(d2 = odd) = 1/2

P(Sum = even) = P(d1=even) * P(d2=even)  + P(d1=odd) * P(d2=odd)
= 1/2*1/2 + 1/2*1/2 = 1/2
Or, we can simply say, sum is either even or odd which are equally likely and hence its probability is 1/2.

we need to find the intersection of events Sum = even and Sum < 6.
P(Sum <6 and Sum= even) = P(Sum = 2 or 4)
= P(d1=d2=1 OR  d1=1, d1=3  OR  d1=3 d2=1  OR  d1=d2=2)
= P(d1=1)*P(d2=1) + P(d1=1)*P(d2=3) + P(d1=3)*P(d2=1) + P(d1=2)*P(d2=2)
= 1/6 *1/6 + 1/6*1/6 + 1/6 */16 + 1/6 * 1/6
= 1/9
Now the conditional probability:
P(Sum < 6 | sum = even) = (1/9) / (1/2) = 2/9
===============================================
b)
Sum = 10 for d1, d2 pair ∈ { 46 , 55, 64 }.  There are 36 pairs possible when two dice are rolled like {11, 12, 13, ,, 21, 22,.., 41,..,44, .. 62,63,64..66 }
P(Sum = 10) = 3 / 36 = 1/12

Roll is doubles =>  d1 d2 pair ∈ { 11, 22, 33, 44, 55, 66 }
P(roll is doubles) = 6/36 = 1/6
The intersection : Sum = even and Roll is doubles = { 55 }
P(Sum = 10 and Roll is doubles) = 1/36
P(Roll is doubles ) = 6/36 = 1/6

P(Sum = 10 | Roll is doubles) = (1/36) / (1/6) = 1/6
============================================
c)
P(Sum = even) = 1/2 --- see part (a)
P(sum < 6 AND sum = even) = 1/9  see part  (a)
P(Sum < 6) = P( {11, 12, 21, 22, 23, 32 } ) = 6/36 = 1/6
P(Sum = even  |  Sum < 6) = (1/9) / (1/6) = 2/3
==========================
d)
P(Sum = 10) = 1/12  see part  (b)
P(Roll is doubles) = 1/6    see part  (b)
P(Sum = 10 AND  roll is doubles) = 1/36    see part  (b)
P(Roll is doubles  | sum = 10) = (1/36) / (1/12) = 1/3
====================
e)
P(Sum > 7) = P( { 26, 35,36,44,45,46,55,56, 66, 65,64,63,62, 54,53 })
= 15/36 = 5/12
P(neither die is a six) = P(1 ≤ d1 ≤ 5)  AND  P(1 ≤ d2 ≤ 5)
= 5/6 * 5/6 = 25/36
P( Sun > 7  AND  Neither die is a six) = P( {35, 44, 45, 55, 54, 53} )
= 6/36 = 1/6
P(Sum > 7 | neither die is a 6) = (1/6) / (25/36) = 6/25
===========================
f)
P(Sum = odd) = 1/2    --  see part (a)
P(at least one die is a six) = P( {16,26,36,46,56,66,65,64,63,62,61 } )
= 11/36
P(sum = odd  AND  at least one die is a six) = P( { 16,36,56,65,63,61 } )
= 6/36 = 1/6
P(Sum is odd  | at least one die is a six)  = (1/6) / (11/36) = 6/11
====================
g)
P(neither die is a six) = 1 - P(at least one die is a six) = 1 - 11/36 = 25/36
P(sum > 7) = 5/12    see part (e)
P(neither die is a six  AND sum > 7) = 1/6    see part  (e)
P(neither die is a six  |  sum > 7) = (1/6) / (5/12) = 2/5
================
h)
P(sum is odd) = 1/2
P(at least one die is a six) = P( {16, 26, 36,46,56, 66, 65,64,63,62,61 } )   = 11/36
P(Sum is odd  AND  at least one die is a six) = P( {16, 36, 56, 65, 63,61} = 6/36 = 1/6
P( at least one die is a six  | Sum is odd ) = (1/6) / (1/2) = 1/3

THANK YOU
c) P(Sum = even) = 1/2 --- see part (a)
P(sum < 6 AND sum = even) = 1/9 see part (a)
P(Sum < 6) = P( {11, 12, 13, 14, 21, 22, 23, 31, 32 } ) = 9/36 = 1/4
P(Sum = even | Sum < 6) = (1/9) / (1/4) = 4/9
sorry for the mistake.
Again, some thing got missed: c) P(Sum = even) = 1/2 --- see part (a)
P(sum < 6 AND sum = even) = 1/9 see part (a)
P(Sum < 6) = P( {11, 12, 13, 14, 21, 22, 23, 31, 32, 41 } ) = 10/36 = 5/18
P(Sum = even | Sum < 6) = (1/9) / (5/18) = 2/5
got the answer right this time.
2015-03-24T22:32:28+05:30

### This Is a Certified Answer

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When two fair dice are rolled, the outcomes are
11,12,13,14,.....,32,33,34,....64,65,66.
number of outcomes when one die is thrown = 6
number of outcomes when two dice are thrown = 6² = 36
So no. of outcomes = 36

a)
outcomes of event E(sum < 6) = (11,12,13,14,21,22,23,31,32,41)
outcomes of event F(sum is even) = (11,13,15,22,24,26,31,33,35,42,44,46,51,53,55,62,64,66)
n(F) = 18
P(F) = 18/36 = 1/2

EΠF = (11,13,22,31)
n(EΠF) = 4
P(EΠF) = 4/36 = 1/9

P(E|F) = P(EΠF) / P(F) = (1/9)/(1/2) = (1×2) / ( 1×9) = 2/9

b)
outcomes of event E(sum = 10) = (46,55,64)
outcomes of event F(roll is doubles) = (11,22,33,44,55,66)
n(F) = 6
P(F) = 6/36 = 1/6

EΠF = (55)
n(EΠF) = 1
P(EΠF) = 1/36

P(E|F) = P(EΠF) / P(F) = (1/36)/(1/6) =(1×6)/(1×36) = 1/6

c)
outcomes of event E(sum is even) = (11,13,15,22,24,26,31,33,35,42,44,46,51,53,55,62,64,66)
outcomes of event F(sum < 6) = (11,12,13,14,21,22,23,31,32,41)
n(F) = 10
P(F) = 10/36 = 5/18

EΠF = (11,13,22,31)
n(EΠF) = 4
P(EΠF) = 4/36 = 1/9

P(E|F) = P(EΠF) / P(F) = (1/9)/(5/18) = (1×18) / ( 5×9) = 2/5

d)
outcomes of event E(roll is doubles) = (11,22,33,44,55,66)
outcomes of event F(sum = 10) = (46,55,64)
n(F) = 3
P(F) = 3/36 = 1/12

EΠF = (55)
n(EΠF) = 1
P(EΠF) = 1/36

P(E|F) = P(EΠF) / P(F) = (1/36)/(1/12) =(1×12)/(1×36) = 1/3

e)
outcomes of event E(sum>7) = (26,35,36,44,45,46,53,54,55,56,62,63,64,65,66)
outcomes of event F(neither die is a 6) =
(11,12,13,14,15,21,22,23,24,25,31,32,33,34,35,41,42,43,44,45,51,52,53,54,5)
n(F) = 25
P(F) = 25/36

EΠF = (35,44,45,53,54,55)
n(EΠF) = 6
P(EΠF) = 6/36 = 1/6

P(E|F) = P(EΠF) / P(F) = (1/6)/(25/36) =(1×36)/(6×25) = 6/25

f)
outcomes of event E(sum is odd) = (12,14,16,21,23,25,32,34,36,41,43,45,52,54,56,61,63,65)
outcomes of event F(atleast one die is a 6) = (16,26,36,46,56,61,62,63,64,65,66,)
n(F) = 11
P(F) = 11/36

EΠF = (16,36,56,61,63,65)
n(EΠF) = 6
P(EΠF) = 6/36 = 1/6

P(E|F) = P(EΠF) / P(F) = (1/6)/(11/36) =(1×36)/(6×11) = 6/11
g)
outcomes of event E(neither die is a 6) =
(11,12,13,14,15,21,22,23,24,25,31,32,33,34,35,41,42,43,44,45,51,52,53,54,55)
outcomes of event F(sum>7) = (26,35,36,44,45,46,53,54,55,56,62,63,64,65,66)
n(F) = 15
P(F) = 15/36 = 5/12

EΠF = (35,44,45,53,54,55)
n(EΠF) = 6
P(EΠF) = 6/36 = 1/6

P(E|F) = P(EΠF) / P(F) = (1/6)/(5/12) =(1×12)/(6×5) = 2/5
h)
outcomes of event E(atleast one die is a 6) = (16,26,36,46,56,61,62,63,64,65,66,)
outcomes of event F(sum is odd) = (12,14,16,21,23,25,32,34,36,41,43,45,52,54,56,61,63,65)
n(F) = 18
P(F) = 18/36 = 1/2

EΠF = (16,36,56,61,63,65)
n(EΠF) = 6
P(EΠF) = 6/36 = 1/6

P(E|F) = P(EΠF) / P(F) = (1/6)/(1/2) =(1×2)/(6×11) = 1/3