Answer
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.
Total number of elementary events = 8
If any of the elementary events HHH, HHT, HTH and THH is an outcome, then we say that the event of “Getting at least two
heads” occurs.
$\therefore $ Number of favorable elementary events = 4
$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$
Hence required probability = $\frac{4}{8} = \frac{1}{2}$
Note:
Probability of an event E is (Number of favorable outcomes to event E) / (Total number of possible outcomes in sample space)
$ \Rightarrow P(E) = \frac{{n(E)}}{{n(S)}}$
In the given problem our event is getting at least two heads while tossing three coins.
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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
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.
`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`