P(x>2x≥5)=P(x>2)P(x≥5∩x>2)
=(65)2⋅61+(65)361+……….∞(65)461+(65)561+.......∞
=1−65(65)2611−65(65)4⋅61=(65)2=3625
A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the conditional probability P(X⩾5∣X>2) is :
Held on 26 Aug 2021 · Verified 6 Jul 2026.
3625
65
3611
216125
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