In a game of darts, the probability of hitting the center of the target is 0.7. what is the probability of getting exactly two successive center hitting in four darts ?
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vipulgoyal
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Anju@Gurome
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Let us assume event of hitting center of the target = H and missing the same = Mvipulgoyal wrote:In a game of darts, the probability of hitting the center of the target is 0.7. what is the probability of getting exactly two successive center hitting in four darts?
So, P(H) = 0.7 and P(M) = (1 - 0.7) = 0.3
Hence, only the following three scenarios are possible...
- HHMM
MHHM
MMHH
Hence, required probability = 3*(0.0441) = 0.1323
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vipulgoyal
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tarunjohri
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srcc25anu
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# ways to select exactly 2 out 4 outcomes to hit the target = 4C2 or 6Anju@Gurome wrote:Let us assume event of hitting center of the target = H and missing the same = Mvipulgoyal wrote:In a game of darts, the probability of hitting the center of the target is 0.7. what is the probability of getting exactly two successive center hitting in four darts?
So, P(H) = 0.7 and P(M) = (1 - 0.7) = 0.3
Hence, only the following three scenarios are possible...Probability of any of the above scenario = P(H)*P(H)*P(M)*P(M) = (0.7)*(0.7)*(0.3)*(0.3) = (0.49)*(0.09) = 0.0441
- HHMM
MHHM
MMHH
Hence, required probability = 3*(0.0441) = 0.1323
they can be listed as
hhmm
hmhm
hmmh
mhhm
mhmh
mmhh
Hence the required probability should be 4C2 * (0.7^2) * (0.3^2) = 0.2646
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Anju@Gurome
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The problem said 'probability of getting exactly two successive center hitting'.srcc25anu wrote:hhmm
hmhm
hmmh
mhhm
mhmh
mmhh
The cases marked with with red are not successive hitting.
@tarunjohri : If that was the case, then the problem should have asked to find out the 'probability of getting at least two successive center hitting'.
Last edited by Anju@Gurome on Sun Apr 21, 2013 12:28 am, edited 1 time in total.
Anju Agarwal
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vipulgoyal
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problem has given excatly 2 successive not at least 2 successive, Anju's first explainatio is correct
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srichinni28
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HHMM,MHHM,MMHH : (0.7)^2 x (0.3)^2 x 3 NOTE : 3 IS NO OF CASES.
+
HHMH, HMHH : (0.7)^3 x (0.3)^1 x 2 NOTE : 2 IS NO OF CASES. IN THESE CASES ALSO EXACTLY 2 SUCCESSIVE CENTER HITTINGS.
ADD BOTH CASES FOR CORRECT ANSWER.
+
HHMH, HMHH : (0.7)^3 x (0.3)^1 x 2 NOTE : 2 IS NO OF CASES. IN THESE CASES ALSO EXACTLY 2 SUCCESSIVE CENTER HITTINGS.
ADD BOTH CASES FOR CORRECT ANSWER.
