Need help as I think there is an error in my book

[Mr2Bits]

A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen to be the secretary and one of the remaining 8 members is to be chosen at random to be the treasurer.

What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?

A – 1/720
B - 1/80
C – 1/10
D- 1/9
E – 1/5

I have worked this out but come up with a different answer than the books explanation and need help reasoning. I will post my answer and the books soon

:

[x2suresh]

A certain club has 10 members, including Harry. One of the 10 members is to be chosen at random to be the president, one of the remaining 9 members is to be chosen to be the secretary and one of the remaining 8 members is to be chosen at random to be the treasurer.

What is the probability that Harry will be either the member chosen to be the secretary or the member chosen to be the treasurer?

A – 1/720
B - 1/80
C – 1/10
D- 1/9
E – 1/5

I have worked this out but come up with a different answer than the books explanation and need help reasoning. I will post my answer and the books soon

:

ways for Harry be sec + ways for Harry to be treasurer
= 9C1*1C1*8C1 + 9C1*8C1*1C1

Total number of ways= 10C1*9C1*8C1


p= (9*8+9*8)/(10*9*8) = 2/10=1/5

E

[Mr2Bits]

E is correct in the book but what I dont get is why they factor him being president in the combination of treasuerer and secretary. Once president is selected it's gone. You cannot elect the president again when the treasurer and secretary positions are being filled.

I came up with

9/10 * 1/9 *1/8 = 1/80 he makes either one.

9/10 being the chance of him not being president

1/9 being the chance he is secretary

1/8 being the chance he is treasurer

Each chance is not independant of the last is it not?

[x2suresh]

E is correct in the book but what I dont get is why they factor him being president in the combination of treasuerer and secretary. Once president is selected it's gone. You cannot elect the president again when the treasurer and secretary positions are being filled.

I came up with

9/10 * 1/9 *1/8 = 1/80 he makes either one.

9/10 being the chance of him not being president

1/9 being the chance he is secretary

1/8 being the chance he is treasurer

Each chance is not independant of the last is it not?

P(AUB) = P(A) + P(B) - P(A ∩ B)

P(A ∩ B) --> ZERO here.. because he can't be secretary and treasurer at the same time.


P(sec or tre)=P(AUB) = P(A) + P(B) = P(sec)+P(tre)
= 9/10*1/9*1/1 + 9/10*8/9*1/8 =2/10=1/5

[kanha81]

Here's how I solved the problem:

Let P(X=P)=1/10, X is someone who is not Harry, P is President
=> P(P')=1-1/10=9/10

lllly, P(H=S)=1/9, H is Harry, S is secretary
=> P(S')=1-1/9=8/9, S' is not Secretary

lllly, P(H=T)=1/8, T is Treasurer
=> P(T')=1-1/8=7/8, T' is not Treasurer

lllly, P(H=NonSelection)=1

Create a Table: Harry can be either Secretary or Treasurer, but not President

PRES SEC TRES
P' S T'
P' S' T

PRES SEC TRES
9/10 1/9 P(Non Selection)=1 (b/c Harry is already elected)
9/10 8/9 1/8

1st row: P(H=S) = (9/10) * (1/9) * 1 = 1/10
2nd row: P(H=T) = (9/10) * (8/9) * (1/8) = 1/10

P(H=S) + P(H=T) = 2(1/10) = 1/5

Hence, [E]

[Mr2Bits]

Thanks all, I get it now, I need to calulate them as two seperate events and not one.