Three unbiased coins are tossed once. Find the probability of getting (i) exactly 2 tails (ii) exactly one tail (iii) at most 2 tails (iv) at least 2

Correct Answer - D `(d)` `P(T)=p`, `P(H)=1-p` `:.` Required probability `=P(T or HHT or HHHHT or ….)` `=p+(1-p)^(2)p+(1-p)^(4)+….` `=(p)/(1-(1-p)^(2))` `=(p)/(2p-p^(2))=(2)/(3)` `impliesp=(1)/(2)`

Correct Answer - `(63)/(100)`

Correct Answer - (i) ` 1/2 (ii) 3/4 (iii) 3/4 `

Correct Answer - `(i) 7/25 (ii) 22/25 (iii) 3/5`

Correct Answer - `(i) 1/6 (ii) 1/3 (iii) 1/2 (iv) 1/2 (v) 1/3 (vi) 1/3`

Correct Answer - (i) 0.57 (ii) 0.43

Correct Answer - (i) 0.28 (ii) 0.4 (iii) 0.32

Correct option is (C) 7/8 If three coins are tossed then possible outcomes are S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} \(\therefore\) Total outcomes is n(S) = 8 Let event E...

Correct option is (C) 1 When an unbiased coin is tossed then possible outcomes are S = {H, T}. \(\therefore\) P(H) = \(\frac12\) & P(T) \(=\frac12\) \(\therefore\) P(H) + P(T) \(=\frac12+\frac12=1\) Hence, sum of probabilities of getting a head...