Question

A biased die has a probability of 1/4 of showing a 5, while the probability of any of 1, 2, 3, 4, or 6 turning up is the same . If three such dice are rolled, what is the probability of getting a sum of atleast 14 without getting a 6 on any die ?
Option 1 : 5/24 Option 2 : 9/160 Option 3 : 1/30 Option 4 : 7/160

Answer 1

ok what have u done so far

Answer 2

Umm dan can i ask you a question

Answer 3

to start off you need to see the prob of getting 14, and the prob of getting 14 with 6s

Answer 4

Like rolling a 5 has a probability of 1/4
but is it right to assume that rolling a 6 has a probability of 3/20

Answer 5

yes

Answer 6

ok that method from what i read so far is fine but

Answer 7

My work:
since the sum should be atleast 14 that means sum could be 14 or 15 or 16 or 17 or 16(6+6+6)
since 6 should not be there on any die we are left with 14 or 15
for 14 and 15
the sum of three dice could be 5+5+4 or 5+5+5
now the probability of getting a 5 is 1/4
so the probability of getting 14 is 1/4 *1/4*1/5
and that of 15 is 1/4 *1/4*1/4
total probability is 1/4 *1/4*1/5+ 1/4 *1/4*1/4

Answer 8

u cannot do that in the case where you have more dices, checking the addition cases is not so easy

Answer 9

what's the probability of getting a 4??

Answer 10

3/4*1/5

Answer 11

there is no other option except adding the sum of 3 dices to get 14 or 15

Answer 12

if this is not the method how will u get 14 or 15
in Question it is asked to add the sum of 3 dices

Answer 13

i think what i did is right
just say may whats the probability of getting a 4 from 1,2,3,4,5,6
should 6 be included to calculate the probability of 4 (in question it is said that 6 should not be used to calculate the sum of 14)

Answer 14

ok just fix prob of 4

Answer 15

that looks fine
with numbers 1-5, you're left with sums of 14 or 15

Answer 16

probability of getting a 4 has nothing to do with the later problem

Answer 17

P(5) = 1/4
P(any other number) = 3/20

Answer 18

the probabbilites are fixed ^
the die wont know your question in advance to change probabilities accordingly

Answer 19

how is it 3/20 can u plz expalin

Answer 20

@ganeshie8

Answer 21

7/160?

Answer 22

@FateRocks8D

Answer 23

The probability of landing on any number from 1-6 is mutually exclusive on a single roll
Meaning you can only spin on 1 ...you cant spin on 1 and 2 simultaneously
Anyways so we know that probability of landing on every single number from 1 to 6 is 1 cus when you roll a die then you will automatically land on a number from 1-6
so therefore we can conclude that
P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1
When we addd the probabilities of every event it obviously equals 1
Now we know that the probability of landing on a 5 is 1/4
So we can substitute P(5)=1/4
So our equations looks like the following

P(1)+P(2)+P(3)+P(4)+(1/4)+P(6)=1

Subtract a 1/4 from both sides we get
P(1)+P(2)+P(3)+P(4)+P(6)=3/4

So we know that the rest all have equal likelihood of landing on its number meaning teh liklihood of landing on 1,2,3,4, and 6 are all the same
hence
P(1)=P(2)=P(3)=P(4)=P(6)=X
So lets represent the probability of landing on any of these 5 numbers as X
So lets replace P(1), P(2) and so forth with X since they all have the same probability

P(1)+P(2)+P(3)+P(4)+P(6)=3/4
X+X+X+X+X=3/4

5X=3/4

\(\frac{5x}{5}=\frac{3}{4} \div 5\)

\(x=\frac{3}{4}*\frac{1}{5}=\frac{3}{20}\)

Answer 24

yeah answer is 7/160

Answer 25

So the probability of landing on 1,2,3,4 and 5 is 3/20

Answer 26

how did you get that answer?