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

22. Three unbiased coins are tossed together. Find the probability of getting (i) two heads. (i) at least two heads.

Answer

Hint- Here, we will be proceeding by analyzing all the possible outcomes when three unbiased coins are tossed together.Given, three unbiased coins are tossed together.The possible cases or outcomes which will arise are given by\[{\text{(H,H,H),(H,H,T),(H,T,H),(T,H,H),(H,T,T),(T,H,T),(T,T,H)}},(T,T,T)\] where H represents head occurring and T represents tail occurring.As we know that the general formula for probability is given as\[{\text{Probability of occurrence of an event}} = \dfrac{{{\text{Total number of favorable outcomes}}}}{{{\text{Total number of possible outcomes}}}}\]Here, the favorable event is the occurrence of two heads when three coins are tossed together.Therefore, favorable cases where two heads occur when three coins are tossed together are \[{\text{(H,H,T),(H,T,H),(T,H,H)}}\].Clearly, Total number of favorable outcomes\[ = 3\] and Total number of possible outcomes\[ = 8\].Therefore, Probability of occurrence of two heads when three unbiased coins are tossed together\[ = \dfrac{3}{8}\].Note- These types of problems are solved with the help of the general formula of probability in which the favorable event is referred to as the event of getting two heads when three unbiased coins are tossed together. Here, all the possible cases which can occur need to be considered.

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`.

Three unbiased coins are tossed. What is the probability of getting a at least two head b at most two heads ctwo heads

Open in App