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  1. Problems
  2. /A coin is tossed three times. Let A be the event "at least two heads", and let B be the event "the…
MediumMathematicsHigh School (11–12)Statistics and Probability › ProbabilityNumerical

A coin is tossed three times. Let A be the event "at least two heads", and let B be the event "the…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A coin is tossed three times. Let AAA be the event "at least two heads", and let BBB be the event "the first toss is a head". List AAA and BBB as sets of strings. Find the number of outcomes in A∩BA \cap BA∩B, in A∪BA \cup BA∪B and in A−BA - BA−B. Then find how many outcomes belong to exactly one of AAA and BBB, and how many belong to neither.

Check your answer

Work it out on paper, then type your number. The answer itself stays hidden.

Need a reminder? The key ideaShowHide

Use n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B)n(A∪B)=n(A)+n(B)−n(A∩B), n(A−B)=n(A)−n(A∩B)n(A - B) = n(A) - n(A \cap B)n(A−B)=n(A)−n(A∩B) and n(A′)=n(S)−n(A)n(A') = n(S) - n(A)n(A′)=n(S)−n(A). For three events add the three singles, take away the three pairs and add the triple. De Morgan: (A∪B)′=A′∩B′(A \cup B)' = A' \cap B'(A∪B)′=A′∩B′.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The sets. A={HHH,HHT,HTH,THH}A = \{HHH, HHT, HTH, THH\}A={HHH,HHT,HTH,THH} and B={HHH,HHT,HTH,HTT}B = \{HHH, HHT, HTH, HTT\}B={HHH,HHT,HTH,HTT}.

Step 2. The overlap. A∩B={HHH,HHT,HTH}A \cap B = \{HHH, HHT, HTH\}A∩B={HHH,HHT,HTH}, so n(A∩B)=3n(A \cap B) = 3n(A∩B)=3.

Step 3. The union. n(A∪B)=4+4−3=5n(A \cup B) = 4 + 4 - 3 = 5n(A∪B)=4+4−3=5. The union is {HHH,HHT,HTH,\{HHH, HHT, HTH,{HHH,HHT,HTH, THH,HTT}THH, HTT\}THH,HTT}.

Step 4. The difference. A−B={THH}A - B = \{THH\}A−B={THH}, so n(A−B)=1n(A - B) = 1n(A−B)=1.

Step 5. Exactly one and neither. Exactly one means A−BA - BA−B or B−AB - AB−A. The set B−AB - AB−A is {HTT}\{HTT\}{HTT}, so 1+1=21 + 1 = 21+1=2. Neither is 8−5=38 - 5 = 38−5=3: THTTHTTHT, TTHTTHTTH and TTTTTTTTT.

Answer: 333, 555, 111, 222 and 333 outcomes.

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