I need help in hypergemetric distrubtion Note: please clear your answer and solv
ID: 3327212 • Letter: I
Question
I need help in hypergemetric distrubtion
Note:
please clear your answer and solve step by step and solve by your hend
Explanation / Answer
In total there are 10 messages of which only 7 have been sent. This is possible in 10C7 ways
= (10!)/(7!)(3!)
= (10 x 9 x 8)/(6)
= 120.
Thus, the denominator for all the probability calculations would be 120…………… (1)
Part (a)
None of the 3 messages is lost => all 3 of 3 and only 4 of 7 are sent. This is possible in
(3C3) x (7C4) ways
= 1 x 35
= 35.
Hence, the required probability = 35/120 = 0.292 ANSWER
Part (b)
Exactly one the 3 messages is lost => only 2 of 3 and 5 of 7 are sent. This is possible in
(3C2) x (7C5) ways
= 3 x 21
= 63.
Hence, the required probability = 63/120 = 0.525 ANSWER
Part (c)
All the 3 messages are lost => none of 3 and all 7 of 7 are sent. This is possible in
(3C0) x (7C7) ways
= 1 x 1
= 1.
Hence, the required probability = 1/120 = 0.0083 ANSWER
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