probability problem

[ganeshiyer]

Originally Posted by suprovo2000

q1. The probability that a graduate student being male is 0.25 and that being female is 0.75. The probability that a male student passes the course is 0.7 and that a female student does it is 0.80. A student selected at random is found to have completed the course. What is the probability that the student is (i) male and (ii) female?



q2. Three members X, Y and Z of a private club have been nominated for the office of the president. The probability the Mr. X will be elected is 0.3, the probability that Mr. Y will be elected is 0.5, and the probability that Mr. Z will be elected is 0.2.Should Mr. X be elected, the probability for an increase in membership fee is 0.8? Should Y or Z be elected, the corresponding probability for an increase in fee is 0.2 and 0.3?If fee has been increased, what is the probability that (i) Mr. X was elected president of the club? (ii) Mr. Y was elected president of the club?

q1

i 0.5

ii 0.5


q2

i 0.3

ii 0.5

am i correct??????????

[clsuresh]

q1

1. 7/31
2. 24/31

Please confirm?

[made_for_iims]

Originally Posted by suprovo2000

q1. The probability that a graduate student being male is 0.25 and that being female is 0.75. The probability that a male student passes the course is 0.7 and that a female student does it is 0.80. A student selected at random is found to have completed the course. What is the probability that the student is (i) male and (ii) female?



q2. Three members X, Y and Z of a private club have been nominated for the office of the president. The probability the Mr. X will be elected is 0.3, the probability that Mr. Y will be elected is 0.5, and the probability that Mr. Z will be elected is 0.2.Should Mr. X be elected, the probability for an increase in membership fee is 0.8? Should Y or Z be elected, the corresponding probability for an increase in fee is 0.2 and 0.3?If fee has been increased, what is the probability that (i) Mr. X was elected president of the club? (ii) Mr. Y was elected president of the club?

Q1........

male prob = (1/4 * 7/10 )/ ( 1/4 * 7/10 + 3/4 * 8/10)

= 7/31

female prob = 24/31


Q2...

prob of x = 0.3*0.8 / 0.3 * 0.8 + 0.5 * 0.2 + 0.2 * 0.2

= 24/38

similarly prob of y = 10/38

[harishkumar]

A fair coin is tossed 10 times.What is the probability of 2 heads not occuring simultaneously?
1> 1/8
2> 1/16
3> 1/32
4> None of these

[Bell-the-Cat]

In all there will be 2^20 cases.



Now, two heads should not occur simultaneaously. We can make the cases as follows:

X T X T X T X T X T X T X T X T X T X T X

Heads can take the place of 'X' above. There are C(11,10) such cases i.e11 such cases.

Hence, the probability is 11/2^20.

Please correct me if I am wrong.

[Apprentice]

Originally Posted by Bell-the-Cat

In all there will be 2^20 cases.



Now, two heads should not occur simultaneaously. We can make the cases as follows:

X T X T X T X T X T X T X T X T X T X T X

Heads can take the place of 'X' above. There are C(11,10) such cases i.e11 such cases.

Hence, the probability is 11/2^20.

Please correct me if I am wrong.

Te ans is :1-(19C2)/2^20...19C2 is choosing 2 consecutive positions for heads to occur...
neways the answer is huge!!!

[anandnat1985]

hii
I think the sample space for this problem should be 2^10.
Please clarify this doubt...
Regards

[Bell-the-Cat]

I got my mistake, I misassumed that there will be only 5 heads, while there can be at min 0 and at max 10 Heads.

So the probability should be

C(11,0)+C(11,1)+C(11,2)+C(11,3)....+C(11,10)=2^11-1

Hence the probability= (2^11-1)/2^20

[Apple]

I think the sample space shud be 2^10.
And my solution is as follows
Case 1: all tails
1 case
CASE 2:
9T 1H
total such cases 10
CASE 3:
8T 2H
XTXTXTXTXTXTXTXTX
we can choose any 2 places frm 9 X = 9C2 =36
CASE 4:
7T 3H
XTXTXTXTXTXTXTX
we can choose any 3 places for H frm 8X = 8C3=56
CASE 5:
6T 4H
by same logic we get 7C4 cases = 35
CASE 6:
5H 5T
by same logic we get 6C5 = 6
so total cases 1 + 10 +36 +56 +35 +6= 144 cases
So answer shd be 144/2^10
pls tell me junta if i m wrong......

[harishkumar]

The above problem which I have posted ( i.e., tossing a coin 10 times) was given in Arun Sharma book. He has not given he solution. But the answer is given as
1/8.