Binomial distribution calculate probability of pikes - is my solution correct?
$begingroup$
In a lake, there are two types of fish: trouts and pikes. Let p = 0.7 be the proportion of trouts in the lake. We pick 20 fish at random with replacement. Let X be the number of trouts.
a) What distribution would correspond to the number of trouts obtained? What is the formula for P(X=k)?
b) What is the probability that we get 18 pikes.
a) I am thinking since we are having sampling with replacement, that the distribution is Binomial. From here on, the formula for $P(k) = binom{n}{k} p^k(1-p)^{n - k}$
b) Since the proportion of trouts is p=0.7, the proportion of pikes in the lake is 0.3. Here k=18 and n=20, so we substitute in the formula above and get: P(18)= 0.000000036.
The result was obtained by using: $P(18) = binom{20}{18} 0.3^{18}0.7^{2}$
This incredible low probability number is rather worrying me and since I don't have the solved solution to compare with a correct answer, I'm asking here if this looks correct or if I messed something up big time?
Thank you in advance for your time and help!
probability statistics binomial-distribution
asked Dec 11 '18 at 21:44
VRTVRT
957
$endgroup$
add a comment |
$begingroup$
In a lake, there are two types of fish: trouts and pikes. Let p = 0.7 be the proportion of trouts in the lake. We pick 20 fish at random with replacement. Let X be the number of trouts.
a) What distribution would correspond to the number of trouts obtained? What is the formula for P(X=k)?
b) What is the probability that we get 18 pikes.
a) I am thinking since we are having sampling with replacement, that the distribution is Binomial. From here on, the formula for $P(k) = binom{n}{k} p^k(1-p)^{n - k}$
b) Since the proportion of trouts is p=0.7, the proportion of pikes in the lake is 0.3. Here k=18 and n=20, so we substitute in the formula above and get: P(18)= 0.000000036.
The result was obtained by using: $P(18) = binom{20}{18} 0.3^{18}0.7^{2}$
This incredible low probability number is rather worrying me and since I don't have the solved solution to compare with a correct answer, I'm asking here if this looks correct or if I messed something up big time?
Thank you in advance for your time and help!
probability statistics binomial-distribution
asked Dec 11 '18 at 21:44
VRTVRT
957
$endgroup$
$begingroup$
How did you obtain your result?
$endgroup$
– callculus
Dec 11 '18 at 21:50
$begingroup$
By P(18) = 20 choose 18 x 0.3 on the 18th x 0.3 on the 2nd. Unsure if formulas work here.
$endgroup$
– VRT
Dec 11 '18 at 21:57
2
$begingroup$
It looks correct to me.
$endgroup$
– DavidPM
Dec 12 '18 at 8:48
$begingroup$
Thank you @DavidPM
$endgroup$
– VRT
Dec 12 '18 at 11:03
add a comment |
$begingroup$
In a lake, there are two types of fish: trouts and pikes. Let p = 0.7 be the proportion of trouts in the lake. We pick 20 fish at random with replacement. Let X be the number of trouts.
a) What distribution would correspond to the number of trouts obtained? What is the formula for P(X=k)?
b) What is the probability that we get 18 pikes.
a) I am thinking since we are having sampling with replacement, that the distribution is Binomial. From here on, the formula for $P(k) = binom{n}{k} p^k(1-p)^{n - k}$
b) Since the proportion of trouts is p=0.7, the proportion of pikes in the lake is 0.3. Here k=18 and n=20, so we substitute in the formula above and get: P(18)= 0.000000036.
The result was obtained by using: $P(18) = binom{20}{18} 0.3^{18}0.7^{2}$
This incredible low probability number is rather worrying me and since I don't have the solved solution to compare with a correct answer, I'm asking here if this looks correct or if I messed something up big time?
Thank you in advance for your time and help!
probability statistics binomial-distribution
asked Dec 11 '18 at 21:44
VRTVRT
957
$endgroup$
In a lake, there are two types of fish: trouts and pikes. Let p = 0.7 be the proportion of trouts in the lake. We pick 20 fish at random with replacement. Let X be the number of trouts.
a) What distribution would correspond to the number of trouts obtained? What is the formula for P(X=k)?
b) What is the probability that we get 18 pikes.
a) I am thinking since we are having sampling with replacement, that the distribution is Binomial. From here on, the formula for $P(k) = binom{n}{k} p^k(1-p)^{n - k}$
b) Since the proportion of trouts is p=0.7, the proportion of pikes in the lake is 0.3. Here k=18 and n=20, so we substitute in the formula above and get: P(18)= 0.000000036.
The result was obtained by using: $P(18) = binom{20}{18} 0.3^{18}0.7^{2}$
This incredible low probability number is rather worrying me and since I don't have the solved solution to compare with a correct answer, I'm asking here if this looks correct or if I messed something up big time?
Thank you in advance for your time and help!
probability statistics binomial-distribution
probability statistics binomial-distribution
asked Dec 11 '18 at 21:44
VRTVRT
957
asked Dec 11 '18 at 21:44
VRTVRT
957
edited Dec 11 '18 at 22:01
VRT
asked Dec 11 '18 at 21:44
VRTVRT
957
asked Dec 11 '18 at 21:44
VRTVRT
957
asked Dec 11 '18 at 21:44
VRTVRT
957
957
$begingroup$
How did you obtain your result?
$endgroup$
– callculus
Dec 11 '18 at 21:50
$begingroup$
By P(18) = 20 choose 18 x 0.3 on the 18th x 0.3 on the 2nd. Unsure if formulas work here.
$endgroup$
– VRT
Dec 11 '18 at 21:57
2
$begingroup$
It looks correct to me.
$endgroup$
– DavidPM
Dec 12 '18 at 8:48
$begingroup$
Thank you @DavidPM
$endgroup$
– VRT
Dec 12 '18 at 11:03
add a comment |
$begingroup$
How did you obtain your result?
$endgroup$
– callculus
Dec 11 '18 at 21:50
$begingroup$
By P(18) = 20 choose 18 x 0.3 on the 18th x 0.3 on the 2nd. Unsure if formulas work here.
$endgroup$
– VRT
Dec 11 '18 at 21:57
2
$begingroup$
It looks correct to me.
$endgroup$
– DavidPM
Dec 12 '18 at 8:48
$begingroup$
Thank you @DavidPM
$endgroup$
– VRT
Dec 12 '18 at 11:03
$begingroup$
How did you obtain your result?
$endgroup$
– callculus
Dec 11 '18 at 21:50
$begingroup$
How did you obtain your result?
$endgroup$
– callculus
Dec 11 '18 at 21:50
$begingroup$
By P(18) = 20 choose 18 x 0.3 on the 18th x 0.3 on the 2nd. Unsure if formulas work here.
$endgroup$
– VRT
Dec 11 '18 at 21:57
$begingroup$
By P(18) = 20 choose 18 x 0.3 on the 18th x 0.3 on the 2nd. Unsure if formulas work here.
$endgroup$
– VRT
Dec 11 '18 at 21:57
2
2
$begingroup$
It looks correct to me.
$endgroup$
– DavidPM
Dec 12 '18 at 8:48
$begingroup$
It looks correct to me.
$endgroup$
– DavidPM
Dec 12 '18 at 8:48
$begingroup$
Thank you @DavidPM
$endgroup$
– VRT
Dec 12 '18 at 11:03
$begingroup$
Thank you @DavidPM
$endgroup$
– VRT
Dec 12 '18 at 11:03
add a comment |
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$begingroup$
How did you obtain your result?
$endgroup$
– callculus
Dec 11 '18 at 21:50
$begingroup$
By P(18) = 20 choose 18 x 0.3 on the 18th x 0.3 on the 2nd. Unsure if formulas work here.
$endgroup$
– VRT
Dec 11 '18 at 21:57
2
$begingroup$
It looks correct to me.
$endgroup$
– DavidPM
Dec 12 '18 at 8:48
$begingroup$
Thank you @DavidPM
$endgroup$
– VRT
Dec 12 '18 at 11:03