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Question : 120 of 150
Marks:
+1,
-0
Solution:
Since,
‌,‌,‌ are probabilities of mutually exclusive events, then
‌0≤‌+‌+‌≤1‌⇒‌‌0≤‌≤1‌⇒‌‌0≤4−p≤4‌⇒‌‌−4≤−p≤0‌⇒‌‌0≤p≤4 . . (i)
‌‌ Also, ‌‌‌0≤‌≤1,0≤‌≤1‌‌ and ‌0≤‌≤1‌⇒‌‌0≤1+4p≤4,0≤1−p≤4and
‌‌0≤1−2p≤2⇒‌‌−‌≤p≤‌,−3≤p≤1and
‌‌−‌≤p≤‌ . . . (ii)
From Eqs. (i) and (ii), we get
0≤p≤‌
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