Solution : What two coins are tossed simultaneously, all possible outcomes are HH, HT, TH, TT.
Their number is 4.
All favourable outcomes are TT,TH,HT. Their number is 3.
`:.` P(getting at most 1 head ) =`3/4`.
Solution : In tossing three coins, the sample space is given gy
`S = {"HHH, HHT, HTH, THH, HTT, THT, TTH, TTT"}.`
And, therefore, n(S) = 8.
(i) Let `E_(1)` = event of getting all heads. Then,
`E_(1) = {HHH}` and, therefore, `n(E_(1)) = 1.`
`therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.`
(ii) Let `E_(2)` = event of getting 2 heads. Then,
`E_(2) = {HHT, HTH, THH}` and, therefore, `n(E_(2)) = 3.`
`therefore` P (getting 2 heads) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/8.`
(iii) Let `E_(3)` = event of getting 1 head. Then,
`E_(3) = {"HTT, THT, TTH" }` and, therefore, `n(E_(3)) = 3.`
`therefore` P (getting 1 head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 3/8.`
(iv) Let `E_(4)` = event of getting at least 1 heads. Then,
`E_(4) = {"HTT, THT, TTH, HHT, HTH, THH, HHH"}` and, therefore, `n(E_(4)) = 7.`
`therefore` P (getting at least 1 head) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 7/8.`
(v) Let `E_(5)` = event of getting at least 2 heads. Then,
`E_(5) = {"HHT, HTH, THH, HHH"}` and, therefore, `n(E_(1)) = 1.`
`therefore` P (getting all heads) `= P(E_(5)) = (n(E_(5)))/(n(S)) = 4/8 = 1/2.`