# Answer in Statistics and Probability for Arche #169832

1.Three coins are tossed : For each discard, the consequence belongs to one of the succeed : { H, T } where H is heads and T is tails. We want to count the total number of tails obtained from those 3 tosses.

Observe that any one of the succeed cases may happen :

1. {TTT} – All the three outcomes are ‘Tail’. Hence, T = 3 in this case.
2. e.g. {TTH} – Any 2 of the 3 outcomes are ‘Tail’ and the remaining one is a ‘Head’ Hence, T = 2 in this case.
3. e.g. {THH} – Any 1 of the 3 outcomes are ‘Tail’ and the remaining two are ‘Head’ Hence, T = 1 in this case.
4. {HHH} – All the three outcomes are ‘Head’. So the number of tails is 0. Hence, T = 0 in this case.

note that, the 4 cases listed above explores all possible outcomes. Hence, the random variable T takes any one value from { 0, 1, 2, 3 }.
2. A mint is flipped four times : For each flip, the result belongs to one of the follow : { H, T } where H is heads and T is tails. We want to count the sum number of tails obtained from those 4 tosses. Observe that any one of the follow cases may happen :

1. {TTTT} – All the four outcomes are ‘Tail’. Hence, T = 4 in this case.
2. e.g. {TTTH} – Any 3 of the 4 outcomes are ‘Tail’ and the remaining one is a ‘Head’ Hence, T = 3 in this case.
3. e.g. {TTHH} – Any 2 of the 4 outcomes are ‘Tail’ and the remaining two are ‘Head’ Hence, T = 2 in this case.
4. e.g. {THHH} – Any 1 of the 4 outcomes are ‘Tail’ and the remaining three are ‘Head’ Hence, T = 1 in this case.
5. {HHHH} – All the four outcomes are ‘Head’. So the number of tails is 0. Hence, T = 0 in this case.

bill that, the 5 cases listed above explores all potential outcomes. Hence, the random variable T takes any one respect from { 0, 1, 2, 3, 4 }.
3. Two balance dice are rolled :

For each roll cube, the result belongs to one of the follow : { 1, 2, 3, 4, 5, 6 } frankincense we can define the following sample space for a two balanced dice : ( 1,1 ) ( 1,2 ) ( 1,3 ) ( 1.4 ) ( 1,5 ) ( 1,6 ) ( 2,1 ) ( 2,2 ) ( 2,3 ) ( 2,4 ) ( 2,5 ) ( 2,6 ) ( 3,1 ) ( 3,2 ) ( 3,3 ) ( 3,4 ) ( 3,5 ) ( 3,6 ) ( 4,1 ) ( 4,2 ) ( 4,3 ) ( 4,4 ) ( 4,5 ) ( 4,6 ) ( 5,1 ) ( 5,2 ) ( 5,3 ) ( 5,4 ) ( 5,5 ) ( 5,6 ) ( 6,1 ) ( 6,2 ) ( 6,3 ) ( 6,4 ) ( 6,5 ) ( 6,6 ) therefore the sum of the issue of dots that will appear S takes any one value from { 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 }
4. Let X be the issue of boys in a family of four children :

Observe that any one of the trace cases may happen :

1. {BBBB} – All the four outcomes are ‘Boy’. Hence, X = 4 in this case.
2. e.g. {BBBG} – Any 3 of the 4 outcomes are ‘Boy’ and the remaining one is a ‘Girl’ Hence, X = 3 in this case.
3. e.g. {BBGG} – Any 2 of the 4 outcomes are ‘Boy’ and the remaining two are ‘Girl’ Hence, X = 2 in this case.
4. e.g. {BGGG} – Any 1 of the 4 outcomes are ‘Boy’ and the remaining three are ‘Girl’ Hence, X = 1 in this case.
5. {GGGG} – All the four outcomes are ‘Girl’. So the number of boys is 0. Hence, X = 0 in this case.

bill that, the 5 cases listed above explores all possible outcomes. Hence, the random variable X takes any one value from { 0, 1, 2, 3, 4 }.

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