binomial probability questions (1 Viewer)

[followme]

hi, i have trouble solving the following questions. could someone help plz. thanks.

Q1) three uniform tetrahedral (triangular pyramid) each has one black face, one white face, one red face and one green face. when tossed onto a table, three faces of each tetrahedral can be seen. Find the probability that:
a) no black face can be seen
b) exactly 2 black faces can be seen
c) at least 2 red faces can be seen
d) 3 white faces and only one green face can be seen.
Ans: 0.015625 , 0.421875, 0.84675, 0.09375

Q2) a dental inspector find that about 20% of childeren of a certain area have tooth decay. Find the probability that out of 6 children examined, only the first, third and fifth have tooth decay.
Ans: 0.02048

Q3) A marksman finds that on the average he hits the target 9 times out of every 10 and scores a bull's eye on the average once every 5 rounds. He fires 4 rounds. Find teh probability that:
a) he hits the target each time
b) he scores at least 2 bull's eye
c) he scores at least 2 bull's eye and he hits that target on each of the 4 rounds.
Ans: 0.6561, 0.1808, 0.1416

 

[followme]

anyone?

 

[gemrism]

How many times is the triangular pyramid tossed?

 

[Riviet]

Q1

a) The event of no black means that the black side of each die is the one face not seen (facing down hidden). This can only occur in one way for each die.

P(no black) = (no. of ways of having all other three colours display)/total possible outcomes
= (1x1x1)/{(4C3)(4C3)(4C3)}
= 0.015625

b) P(2 black) = (no. of ways of choosing 2 die out of 3 which show the black face)/(total outcomes)
=(3C2.3C2.3C2)/{(4C3)(4C3)(4C3)}
=0.421875
Reasoning: we have 2 black faces shown for 2 die and there are 3C2 ways for each of these 2 die in choosing the 2 remaining colours (out of 3 available) for the 2 other faces that are visible. The third 3C2 is the number of ways of choosing 2 dice out of 3 that have the black face visible.

 

[renya]

1.a) prob the black face cant be seen=1/4=0.25
prob no black=0.25 x 0.25 x 0.25= 0.015625

b)prob can see black=3/4=0.75
prob exactly 2 black= 0.75 x 0.75 x 0.25 x3=0.421875

c)prob at least 2 red=prob exactly 2 red + prob 3 red
=0.421875 + .75x.75x.75=0.84375 (possibly error in answers)

d)3 white faces and only one green
prob 3 white=.75x.75x.75=0.421875
AND
prob only one green=(2/3)x(1/3)x(1/3)x3
[2/3 now instead of 3/4 because one of the seeable faces is white]
prob 3 white one green= 0.421875 x (2/3)x(1/3)x(1/3)x3=0.09375

 

[Riviet]

Q2) a dental inspector find that about 20% of childeren of a certain area have tooth decay. Find the probability that out of 6 children examined, only the first, third and fifth have tooth decay.
Ans: 0.02048

20%=1/5 have tooth decay => 4/5 don't have tooth decay
P(1st, 3rd and 5th have tooth decay) = p(1st has tooth decay).p(2nd doesn't have tooth decay).p(3rd has tooth decay).p(4th doesn't have tooth decay).p(5th has tooth decay).p(6th doesn't have tooth decay)
=1/5.4/5.1/5.4/5.1/5.4/5
=0.02048

 

[pLuvia]

3)
a) P(hits target)=4C4*(0.9)⁴=0.6561
b) P(gets a bullseye at least twice)=1-4C0*(1/5)⁰(4/5)⁴-4C1*(1/5)¹(4/5)³=0.1808

 

[followme]

Thanks people!

20%=1/5 have tooth decay => 4/5 don't have tooth decay
P(1st, 3rd and 5th have tooth decay) = p(1st has tooth decay).p(2nd doesn't have tooth decay).p(3rd has tooth decay).p(4th doesn't have tooth decay).p(5th has tooth decay).p(6th doesn't have tooth decay)
=1/5.4/5.1/5.4/5.1/5.4/5
=0.02048

(0.2)^3 x (0.8)^3 =/= 0.02048. That's what i did too, or.. could the answer be wrong?

and another question: does AND always mean multiply?

 

[Riviet]

Thanks people!


(0.2)^3 x (0.8)^3 =/= 0.02048. That's what i did too, or.. could the answer be wrong?

and another question: does AND always mean multiply?

The answer's probably wrong, and "and" usually means to multiply, but check with the context of the question.

 

[followme]

and "and" usually means to multiply, but check with the context of the question.

okay thanks. eg for Q3 c) 0.6561x0.1808 =/= 0.1416 even though there is an 'AND'.
lol, dun tell the the answer is wrong again.

 

[Riviet]

I think it might have been a typo with the textbook, it's just a decimal point off.

 

[followme]

okay. thank u all very much!