|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Dec 2007 20:48:41 IST
|
|
|
Q.A,B,C are aimimg to shoot a balloon.A will succeed 4 times out of 5 attempts.The chance of B to shoot the balloon is 3 out of 4 and that of C is 2 out of 3.if 3 aim the balloon simultaneously , the probability that at least 2 of them hitting the balloon is?
|
|
|
|
|
|
|
|
possible cases are
1--> no one hits balloon
2---> 1 of them hits balloon
3--> 2 of them hit balloon
4--> all of them hit the target
probability that no one hits balloon = P(z) = 1/5 * 1/4 * 1/3 = 1/60
probability that only one of them hits the balloon =
P(x) = 4/5 * 1/4 * 1/3 + 1/5*3/4*1/3 + 1/5*1/4*2/3 = 9/60
now proability that atleast two of them hit balloon = 1- P(z) - P(x)
= 1- 10/60 = 5/6
|
The heights which great men reached and kept, Were not attained by sudden flight, They, whilst their companions slept, Were toiling upwards in the night.... |
this reply: 5 points
(with 1 
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
4 Dec 2007 22:43:20 IST
|
|
|
A=4/5
B=3/4
C=2/3
Required Probab -
ABC(All Three) + ABC'(a/b but not c) + AB'C + A'BC
[ 24 + 12 + 8 + 6 ] / 60
50/60
5/6
^^ Is Another Way of doing it........Similar To Nishant's Method Though
|
Sometimes One Dream Is Enough To Light Up The Entire October Sky....
First Year Mechanical Engineering
Veermata Jijabai Technological Institute |
this reply: 5 points
(with 1
in 1 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
