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Re: A certain experiment has three mutually exclusive possible outcomes. P [#permalink]
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It is given that a certain experiment has three mutually exclusive possible outcomes where, ๐‘ƒ (First outcome) = x+1/5 and ๐‘ƒ (second outcome) = 3/10-x


Let, ๐‘ƒ (third outcome) = ๐‘ƒ(๐ถ)

We need to find the probability of the third outcome. Since it is given that the event has only three possible outcomes that means events are mutually exclusive, hence ๐‘ƒ(๐ด) + ๐‘ƒ(๐ต) + ๐‘ƒ(๐ถ) = 1, where ๐‘ƒ(๐ด),๐‘ƒ(๐ต) and ๐‘ƒ(๐ถ) are the probabilities of the three events.

As, ๐‘ƒ(๐ด) = ๐‘ฅ +1/5, ๐‘ƒ(๐ต) =3/10โˆ’ ๐‘ฅ
Hence, using ๐‘ƒ(๐ด) + ๐‘ƒ(๐ต) + ๐‘ƒ(๐ถ) = 1

๐‘‚๐‘Ÿ, ๐‘ฅ +1/5+3/10 โˆ’ ๐‘ฅ + ๐‘ƒ(๐ถ) = 1

๐‘‚๐‘Ÿ, ๐‘ƒ(๐ถ) = 1 โˆ’1/5โˆ’3/10
๐‘‚๐‘Ÿ, ๐‘ƒ(๐ถ) = 1 โˆ’1/2=1/2
๐‘‚๐‘Ÿ, ๐‘ƒ(๐ถ) =1/2
Ans. (C)
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Re: A certain experiment has three mutually exclusive possible outcomes. P [#permalink]
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