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Verified Hint : When we toss an unbiased coin we get either a head or tail. Here we are throwing three such coins. We are going to proceed with the thought of getting head and tail sequence on three coins. Elementary events associated with the random experiment of tossing three coins are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT. Note: Open in App Solution When 3 unbaised coins are tossed, the total number of events = 8 They are {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} (1) Number of favourable events (getting at least two heads) = 4 Probability of getting at least two heads = 4/8 = 1/2 (2) At most two heads (adsbygoogle = window.adsbygoogle || []).push({}); That is no head or one head or 2 heads Probability of getting at most two heads = 7/8 3) Two head = 3 (HHT,HTH,THH) Probability of getting no head = 3/8. (adsbygoogle = window.adsbygoogle || []).push({});`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`. Solution : `3` unbiased coins are tossed together <br> sample space `S` <br> `={HHH,HHT,HTH,THH,TTH,THT,HTT,TTT}` <br> no. of events, n(s)=`8` <br> (1) For getting one head, we will have the event space, <br> `S_1={TTH,THT,HTT}` <br> Number of events,`n(s_1)=3` <br> Thus, probability of getting one head <br> `P(S_1)=(n(S_1))/(n(S))` <br> `=3/8` <br> (2) For getting two heads, we will have the event space, <br> `S_2={HHT,HTH,THH}` <br> Number of events, `n(S_2)=3` <br> Thus, probability of getting two heads, <br> `P(S_2)=(n(s_2))/(n(S))` <br> `=3/8` <br> (3) For getting all heads, we will have the sample space <br> `S_3={HHH}` <br> Number of events,`n(s_3)=1` <br> Thus probability of getting all heads <br> `P(S_3)=(n(S_3))/(n(S))` <br> `1/8` <br> (4) For getting T least two heads, we will have the sample space, <br> `S_4={HHH,HHT,HTH,THH}` <br> Number of events,`n(s_4)=4` <br> Thus, prbability of at least two heads <br> `P(S_4)=(n(s_4)/n(S)` <br> `4/8`=`1/2` What is the probability of getting 2 heads when 3 unbiased coins are tossed?<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`.
What is the probability of getting at least 2 tails if 3 unbiased coins are tossed?P ( getting at least two tails) =84=21.
What is the probability of getting one head if three unbiased coins are tossed?A coin is tossed 3 times. The probability of getting at least one head is 7/8.
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