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Probablity

by vipulgoyal » Tue Mar 19, 2013 2:31 am
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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by Anju@Gurome » Tue Mar 19, 2013 2:56 am
vipulgoyal 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?
Let us assume event of hitting center of the target = H and missing the same = M
So, P(H) = 0.7 and P(M) = (1 - 0.7) = 0.3

Hence, only the following three scenarios are possible...
  • HHMM
    MHHM
    MMHH
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

Hence, required probability = 3*(0.0441) = 0.1323
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by vipulgoyal » Tue Mar 19, 2013 3:12 am
Thanks very well explained
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by tarunjohri » Sat Apr 20, 2013 7:00 am
HHMM
MHHM
MMHH

What about

HHMH
HMHH??
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by srcc25anu » Sat Apr 20, 2013 12:05 pm
Anju@Gurome wrote:
vipulgoyal 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?
Let us assume event of hitting center of the target = H and missing the same = M
So, P(H) = 0.7 and P(M) = (1 - 0.7) = 0.3

Hence, only the following three scenarios are possible...
  • HHMM
    MHHM
    MMHH
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

Hence, required probability = 3*(0.0441) = 0.1323
# ways to select exactly 2 out 4 outcomes to hit the target = 4C2 or 6

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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by Anju@Gurome » Sat Apr 20, 2013 7:39 pm
srcc25anu wrote:hhmm
hmhm
hmmh
mhhm
mhmh
mmhh
The problem said 'probability of getting exactly two successive center hitting'.
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.
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by vipulgoyal » Sun Apr 21, 2013 12:14 am
problem has given excatly 2 successive not at least 2 successive, Anju's first explainatio is correct
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by srichinni28 » Fri Nov 08, 2013 4:43 am
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.
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