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29 Nov 2021, 03:45
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
14% (02:00) correct
86%
(01:07)
wrong
based on 64
sessions
History
Date
Time
Result
Not Attempted Yet
Two airplanes A and B bomb a target in succession. The probabilities of A and B scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. Find the probability that the second plane hits the target.
A. 0.06
B. 0.14
C. 0.32
D. 0.7
E. 0.5
A. 0.06
B. 0.14
C. 0.32
D. 0.7
E. 0.5
11 Dec 2021, 03:34
Kudos
Bookmarks
I am getting B as the answer, could anyone please explain how C is the answer.
TIA
TIA
11 Dec 2021, 11:39
Kudos
Bookmarks
The difficulty with this problem relies significantly on its opaque wording.
It should be read as each plane will alternate bombing with the other until one scores a hit, and what is the probability that B will be the successful bomber.
The probability of B being the successful bomber requires A to continually miss and permits B to miss until it, at infinity, eventually hits.
Let S = the probability of B hitting.
S= (.7*.2)+(.7*.8*.7*.2)+(.7*.8*.7*.8*.7*.2)+...
Where .7 and .8 are probability of A and B missing.
This series is infinite.
Simplifying
S= (.7*.2)+(.7^2*.8*.2)+(.7^3*.8^2*.2)+...
If we divide both sides by (.7*.8)
S/(.7*.8) = (.2/.8)+(.7*.2)+(.7^2*.8*.2)...
Notice the right side equals S, except for the added (.2/.8), so
S/(.7*.8) = S+(1/4)
(S/.56)-S = 1/4
S((1/.56)-1)= 1/4
S((1-.56)/.56)= 1/4
S(.44/.56)= 1/4
S = (1/4)*(56/44)
S= 14/44= 7/22= .318> C
Posted from my mobile device
It should be read as each plane will alternate bombing with the other until one scores a hit, and what is the probability that B will be the successful bomber.
The probability of B being the successful bomber requires A to continually miss and permits B to miss until it, at infinity, eventually hits.
Let S = the probability of B hitting.
S= (.7*.2)+(.7*.8*.7*.2)+(.7*.8*.7*.8*.7*.2)+...
Where .7 and .8 are probability of A and B missing.
This series is infinite.
Simplifying
S= (.7*.2)+(.7^2*.8*.2)+(.7^3*.8^2*.2)+...
If we divide both sides by (.7*.8)
S/(.7*.8) = (.2/.8)+(.7*.2)+(.7^2*.8*.2)...
Notice the right side equals S, except for the added (.2/.8), so
S/(.7*.8) = S+(1/4)
(S/.56)-S = 1/4
S((1/.56)-1)= 1/4
S((1-.56)/.56)= 1/4
S(.44/.56)= 1/4
S = (1/4)*(56/44)
S= 14/44= 7/22= .318> C
Posted from my mobile device
