Three unbiased coins are tossed together what is the probability of getting almost two heads

Q:

X attempts 100 questions and gets 340 marks. If for every correct answer is 4 marks and wrong answer is negative one mark, then the number of questions wrongly answered by Mr. X is:

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Q:

If the standard deviation of 0, 1, 2, 3 ......... 9 is K, then the standard deviation of 10, 11, 12, 13 ........... 19 will be:

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Q:

Find the range of the data 2, 1, 2, 3, 5, 4, 7, 3, 5, 2, 4.

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4 2804

Q:

Find the median, mode and mean of 9, 5, 8, 9, 9, 7, 8, 9, 8.

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Q:

Find the range and mode of the data 17, 18, 28, 19, 16, 18, 17, 29, 18

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7 2623

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.

Answer

Verified

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.

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`

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