The probability of 2 "times" is 0,25
\(x^2=0,25\)
\(x=0,5\)
x= probability of missing/hitting the target

The probability that he will miss the next shot is 0,5.
"The streak" is not relevant.If I throw a coin the probability of getting one "side" is the same regardless of what I have obtained in the previous shots.

Correct me if I am wrong.

Probablity of B can shoot a target 2/2 is 0.25 i.e. the probablity that he will shot both the times is 1/4.
And the probality that he will score either 1 or 0 (miss 1 or both) is 3/4.
Hence the probablity of miss the target should be 3/4 i.e. 0.75.
Answer is C.

OA is B and the above explanation is correct.
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Suppose there are total 'x' shots.
B shoots 2 out of 2 times. means out of x shots (x>2) B shots 2 out of 2 ----> B shots at least 2 out of x.
therefore, 2/x = 0.25 (given)

the target will be missed by B immediately after such two shots : this means he can shot just twice ...the third shot cannot happen which means he missed (x-2) shots.
thus, the probabilty of missing just after 2 shots is (x-2)/x.
(x-2)/x = 1 - 2/x
= 1 - 0.25
= 0.75

Answer : C

Please correct me.

Frankly I don't understand your reasoning above. Anyway:

The probability that B can shoot a target two out of two times is 0.25 --> if the probability of success is p, then the probability of two successes in a row is p*p, we are told that p*p=0.25 --> p=0.5.

Now, the third shoot is independent from the first two, thus the probability of success with third try is p=0.5 and the probability of missing in third try is 1-p=0.5.

Answer: B.
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hey bunuel ....i still cant understand the method:(
why v r considering p=0.25 ... while in ques its clearly mentioned that success for two shots is 0.25 ....

Please read carefully, it's \(p*p=0.25\) --> \(p=0.5\).
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ok bunuel..thanks a lot :D

There are only 2 possibilietes: hit or no hit. Hence for the first shot the probability of hitting the target is 1/2. For the second shot also 1/2. Thus, the probability of hitting both is 1/2*1/2 = 1/4. Not hitting the next target is again 1/2. Hence B.

Hello Bunuel,

How is the third shot independent of the first two shots ? The question states that "What is the probability that the target will be missed by B immediately after such two shots".
I assumed this to be a conditional probability. That is, we have to consider only those failures which follow two successes.

Appreciate your thoughts on this ! Thank you

Agree that the wording is not perfect. But "missed by B immediately after such two shots" means that these two shoots have already been made.
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probability of at least 1 miss in the 2 hits is 0.75. (1-0.25). so (miss,hit)(hit,miss) (miss miss) is 0.75. in these 2 out of three is miss in the first attempt(which replicates for 3rd attempt). hence
(2/3)*0.75=0.5.

Regds
Siva

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