Question

did I get my answers right?
An expert shot hits the target 95% of the time. What are the probabilities that he will miss the target:
a.for the first time on the fifteenth shot?
b.at least one time during fifteen shots?
c.for the second time on the eighteenth shot?

a) P(X=0;14, 1/20) * P(X=1;1, 1/19) ≈ 0.3593 * 0.05 ≈ 0.0180

b) P(X=1;15, 1/20) ≈ 0.3658

c) P(X=1;1, 1/20) * P(X=1;7, 1/20) ≈ 0.05 * 0.2708 ≈ 0.0349

Answer 1

How did you obtain 0.3593 in a)?

Answer 2

susing the formula for binomial distributions[ n!/(x!(n-x)!) ]* p^x *q^(n-x)

Answer 3

Okay

Answer 4

But, in P(X=1;1, 1/19) what does 1/19 mean ?

Answer 5

I think p represents probability of success which is 95/100 i.e. 19/20 and q represents probability of failure which is 5/100 i.e. 1/20 . Is this right ?

Answer 6

yes that's right

Answer 7

and x is the # of 'successes'

Answer 8

You should then have used directly this

for (a) required answer is of getting P(X = 14) in 14 trials and then multiplying with P(X=0) i.e. no success in 1 trial (which is the 15th one).

Answer 9

Here by P(X=k) I mean probability of getting exactly k successes in n trials.

Answer 10

So, using my method of representation:

(A) P(X=14;14)*P(X=0;1)

Answer 11

I was Using P(X=0; 14, 1/20) because I thought that the 'success' here was not hitting a target

Answer 12

So for (A) the answer should be:

\[\large{\left(\begin{matrix}14 \\14\end{matrix}\right)*\left( \cfrac{ 19 }{ 20 } \right)^{14}*\left( \cfrac{ 1 }{ 20 } \right)^{0}}\]

Answer 13

or is that just semantics?

Answer 14

Well you can take success for that also, but your main mistake was in this

\(\color{blue}{\text{Originally Posted by}}\) @pblaesre

a) P(X=0;14, 1/20) * \(\large{P(X=1;1, 1/19)}\) ≈ 0.3593 * 0.05 ≈ 0.0180

\(\color{blue}{\text{End of Quote}}\)

Answer 15

In the enlarged part, the probability for failure should have been 19/20

Answer 16

yes I see that thanks,

Answer 17

For (B) it should be :

Req. probability P = 1 - P(X=0;15)

Remember here I have used your notations.

Answer 18

It means the required probability should be 1 - probability of not missing any target i 15 trials.

Answer 19

shouldn't it be:
P = 1 - P(X=1;15)
since it is
"b.at least one time during fifteen shots?"

Answer 20

At least one's conjugate is not even one

Answer 21

So, it should be

P = 1 - P(X=0;15)

Answer 22

Did you get till this step ?

Answer 23

Sorry I'm still mentally working on the first one, I want to make sure I have it right

Answer 24

Okay

Answer 25

I'm a bit slow haha

Answer 26

Its okay

Answer 27

For missing first time on 15th shot you have to ensure that no shot is missed in first 14 trials and then the shot must miss on 15th trial.

Right ?

Answer 28

I think I got that

Answer 29

(A) [P(X=14;14, 19/20)*P(X=0;1, 19/20)]
= [14P14 (19/20)^(14)(1/20)^(0) * 0P1 (19/20)^(0)(1/20)^(1)]
≈ 0.0244

Answer 30

Its actually 14C14 and 1C0

Answer 31

so the first 14 are successes and the last on is a failure

Answer 32

Yes if you mean by success not missing and by failure - missing

Answer 33

yes

Answer 34

Good.

Answer 35

the book i'm reading says it's a type of permutaion ( a permutation of similar objects) butit looks like a combination

Answer 36

SO your answer for (A) would be :

\[\large{\left(\begin{matrix}14 \\ 14\end{matrix}\right)*(\cfrac{19}{20})^{14} * \left(\begin{matrix}1 \\ 0\end{matrix}\right)*(\cfrac{1}{20})^{1}}\]

Answer 37

Yes it is a combination

Answer 38

Here, \[\left(\begin{matrix}n \\ r\end{matrix}\right) = \cfrac{n!}{(r!)(n-r)!}\]

Answer 39

yup that's what I got

Answer 40

Good

Answer 41

Then move on to (B).

Answer 42

\[\left(\begin{matrix}15 \\ 0\end{matrix}\right) \left( \frac{ 19 }{ 20 } \right)^{15}\]

Answer 43

is that even close?

Answer 44

1-that =0.5367

Answer 45

Great work @pblaesre You really grabbed the concept.

Answer 46

thanks :)

Answer 47

Lets move on to (C).

Answer 48

Shall we ?

Answer 49

Is it okay if I try and check with you if i'm right?

Answer 50

That's fine by me.

Answer 51

GO ahead give it a try

Answer 52

\[\left(\begin{matrix}17 \\ 16\end{matrix}\right)\left(\begin{matrix}\frac{ 19 }{ 20 } \\ \end{matrix}\right)^{16}\left(\begin{matrix}\frac{ 1 }{ 20 } \\ \end{matrix}\right)^{1} \times \left(\begin{matrix}1 \\0\end{matrix}\right)\left(\begin{matrix}\frac{ 19 }{ 20 } \\ \end{matrix}\right)^{0}\left(\begin{matrix}\frac{ 1 }{ 20 } \\ \end{matrix}\right)^{1}\]

from
P( X=16;17) * P(X=0;1)

Answer 53

Great work indeed @pblaesre , great work. You got that right, Perfect answer.

Answer 54

thanks! this was really confusing me at first.

Answer 55

No problem. But, since I can't give more than one medal you will have to get satisfied with my words of appreciation only. Seriously, you have done great work. Keep this hard work and you will master probability in no time.

Best of luck

Vishwesh