Math Help - probability....

1. probability....

Hello,
please try to solve this question.Thanks.
Question no:1
The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

(a)at least 10 survive
(b)from 3 to 8 survive
(c)exactly 5 survive

2. Originally Posted by m777
Hello,
please try to solve this question.Thanks.
Question no:1
The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

(a)at least 10 survive
(b)from 3 to 8 survive
(c)exactly 5 survive
The number of survivors has a binomial distribution $B(15, 0.4)$, so:

$
p(n\ survivors) = b(n;15,0.4)=\frac{15!}{n!(15-n)!}(0.4)^n(1-0.4)^{15-n}
$

(a) $p(10\ or\ more\ survivors) = \sum_{n=10}^{15}b(n;15,0.4)$

(b) $p(3\ to\ 8\ survivors) = \sum_{n=3}^{8}b(n;15,0.4)$

(c) $p(5\ survivors) = b(5;15,0.4)$

RonL

3. Hello, m777!

Do they really expect us to crank out these answers?

The probability that a patient recovers from a rare blood disease is 0.4.
If 15 people are known to have contracted this disease, what is the probability that:

(a) at least 10 survive
(b) from 3 to 8 survive
(c) exactly 5 survive

(a) "At least 10" means: 10 or 11 or 12 or 13 or 14 or 15.

$P)10) \:=\: {15\choose 10}(0.4)^{10}(0.6)^5$
$P(11) \:=\: {15\choose11}(0.4)^{11}(0.6)^4$
$P(12) \:=\: {15\choose12}(0.4)^{12}(0.6)^3$
$P(13) \:=\: {15\choose13}(0.4)^{13}(0.6)^2$
$P(14) \:=\: {15\choose14}(0.4)^{14}(0.6)^1$
$P(15) \:=\: {15\choose15}(0.4)^{15}(0.6)^0$

. . Now add them up . . .

(b) "From 3 to 8" means: 3 or 4 or 5 or 6 or 7 or 8.

$P(3) \:=\:{15\choose3}(0.4)^3(0.6)^{12}$
$P(4) \:=\:{15\choose4}(0.4)^4(0.6)^{11}$
$P(5) \:=\:{15\choose5}(0.4)^5(0.6)^{10}$
$P(6) \:=\:{15\choose6}(0.4)^6(0.6)^9$
. $\vdots\qquad\qquad\qquad\vdots$

. . You get the idea . . .

(c) Exactly 5 . . . We did this one in part (b).

$P(5)\:=\:\frac{15!}{5!10!}(0.4)^5(0.6)^{10} \:=\:0.0612979282$

4. Originally Posted by Soroban
Hello, m777!

Do they really expect us to crank out these answers?

(a) "At least 10" means: 10 or 11 or 12 or 13 or 14 or 15.

$P)10) \:=\: {15\choose 10}(0.4)^{10}(0.6)^5$
$P(11) \:=\: {15\choose11}(0.4)^{11}(0.6)^4$
$P(12) \:=\: {15\choose12}(0.4)^{12}(0.6)^3$
$P(13) \:=\: {15\choose13}(0.4)^{13}(0.6)^2$
$P(14) \:=\: {15\choose14}(0.4)^{14}(0.6)^1$
$P(15) \:=\: {15\choose15}(0.4)^{15}(0.6)^0$

. . Now add them up . . .

(b) "From 3 to 8" means: 3 or 4 or 5 or 6 or 7 or 8.

$P(3) \:=\:{15\choose3}(0.4)^3(0.6)^{12}$
$P(4) \:=\:{15\choose4}(0.4)^4(0.6)^{11}$
$P(5) \:=\:{15\choose5}(0.4)^5(0.6)^{10}$
$P(6) \:=\:{15\choose6}(0.4)^6(0.6)^9$
. $\vdots\qquad\qquad\qquad\vdots$

. . You get the idea . . .

(c) Exactly 5 . . . We did this one in part (b).

$P(5)\:=\:\frac{15!}{5!10!}(0.4)^5(0.6)^{10} \:=\:0.0612979282$

My answer is 0.015603175...i using this formula
10C5 x (0.4)^5 x (0.6)^10...

5. None of them will survive. In 200 years, they will all die...I'm 99% sure of that.

Edit: I've checked and double checked my data and it appears I had made a mistake... it should be 87.3%.

6. Originally Posted by rualin
None of them will survive. In 200 years, they will all die...I'm 99% sure of that.
That 99% should by tradition be 87.3% (since as all professionals know 87.3% of all statistics are made up)

RonL