Three unbiased coins are tossed together find the probability of getting exactly two heads

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Text Solution

Solution : In tossing three coins, the sample space is given gy <br> `S = {"HHH, HHT, HTH, THH, HTT, THT, TTH, TTT"}.` <br> And, therefore, n(S) = 8. <br> (i) Let `E_(1)` = event of getting all heads. Then, <br> `E_(1) = {HHH}` and, therefore, `n(E_(1)) = 1.` <br> `therefore` P (getting all heads) `= P(E_(1)) = (n(E_(1)))/(n(S)) = 1/8.` <br> (ii) Let `E_(2)` = event of getting 2 heads. Then, <br> `E_(2) = {HHT, HTH, THH}` and, therefore, `n(E_(2)) = 3.` <br> `therefore` P (getting 2 heads) `= P(E_(2)) = (n(E_(2)))/(n(S)) = 3/8.` <br> (iii) Let `E_(3)` = event of getting 1 head. Then, <br> `E_(3) = {"HTT, THT, TTH" }` and, therefore, `n(E_(3)) = 3.` <br> `therefore` P (getting 1 head) `= P(E_(3)) = (n(E_(3)))/(n(S)) = 3/8.` <br> (iv) Let `E_(4)` = event of getting at least 1 heads. Then, <br> `E_(4) = {"HTT, THT, TTH, HHT, HTH, THH, HHH"}` and, therefore, `n(E_(4)) = 7.` <br> `therefore` P (getting at least 1 head) `= P(E_(4)) = (n(E_(4)))/(n(S)) = 7/8.` <br> (v) Let `E_(5)` = event of getting at least 2 heads. Then, <br> `E_(5) = {"HHT, HTH, THH, HHH"}` and, therefore, `n(E_(1)) = 1.` <br> `therefore` P (getting all heads) `= P(E_(5)) = (n(E_(5)))/(n(S)) = 4/8 = 1/2.`

Text Solution

`1/8` `7/8` `3/8``1/4`

Answer : B

Solution : When 3 coins are tossed simultaneously, all possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT. <br> Total number of possible outcomes = 8. <br> (i) Let `E_(1)` be the event of getting exactly 2 heads. <br> Then, the favourable outcomes are HHT, HTH, THH. <br> Number of favourable outcomes = 3. <br> ` :. ` P(getting exactly 2 heads) = ` P(E_(1)) = 3/8`. <br> (ii) Let `E_(2)` be the event of getting at least 2 heads. <br> Then, `E_(2)` is the event of getting 2 or 3 heads. <br> So, the favourable outcomes are <br> HHT, HTH, THH, HHH. <br> Number of favourable outcomes = 4. <br> `:. ` P(getting at least 2 heads) = `P(E_(2)) = 4/8 = 1/2`. <br> (iii) Let `E_(3)` be the event of getting at most 2 heads. <br> Then, `E_(3)` is the event of getting 0 or 1 head or 2 heads. <br> So, the favourable outcomes are <br> TTT, HTT, THT, TTH, HHT, HTH, THH. <br> Number of favourable outcomes = 7. <br> `:. ` P(getting at most 2 heads ) = `P(E_(3)) = 7/8`.