Question

Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”. Which events are
(i) Mutually exclusive? (ii) Simple? (iii) Compound?
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Answer 1

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$\Large\fbox{\color{purple}{QUESTION}}$

Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”, C denote the event” three tails show and D denote the event ‘a head shows on the first coin”. Which events are

(i) Mutually exclusive?

(ii) Simple?

(iii) Compound?

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$\bf\Huge\red{\mid{\overline{\underline{ ANSWER }}}\mid }$

➡️Since either coin can turn up Head (H) or Tail (T), are the possible outcomes.

➡️But, now three coins are tossed once so the possible sample space contains,

➡️S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTH}

➡️Now,

➡️A: ‘three heads’

➡️A= (HHH)

➡️B: “two heads and one tail”

➡️B= (HHT, THH, HTH)

➡️C: ‘three tails’

➡️C= (TTT)

➡️D: a head shows on the first coin

➡️D= (HHH, HHT, HTH, HTT)

✴➡️(i) Mutually exclusive

➡️A ∩ B = (HHH) ∩ (HHT, THH, HTH)

➡️= φ

➡️Therefore, A and C are mutually exclusive.

➡️A ∩ C = (HHH) ∩ (TTT)

➡️= φ

➡️There, A and C are mutually exclusive.

➡️A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT)

➡️= (HHH)

➡️A ∩ D ≠ φ

➡️So they are not mutually exclusive

➡️B ∩ C = (HHT, HTH, THH) ∩ (TTT)

➡️= φ

➡️Since there is no common element in B & C, so they are mutually exclusive.

➡️B ∩ D = (HHT, THH, HTH) ∩ (HHH, HHT, HTH, HTT)

➡️= (HHT, HTH)

➡️B ∩ D ≠ φ

➡️Since there are common elements in B & D,

➡️So, they not mutually exclusive.

➡️C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT)

➡️= φ

➡️Since there is no common element in C & D,

➡️So they are not mutually exclusive.

✴➡️(ii) Simple event

➡️If an event has only one sample point of a sample space, it is called a simple (or elementary) event.

➡️A = (HHH)

➡️C = (TTT)

➡️Both A & C have only one element,

➡️so they are simple events.

✴➡️(iii) Compound events

➡️If an event has more than one sample point, it is called a Compound event

➡️B= (HHT, HTH, THH)

➡️D= (HHH, HHT, HTH, HTT)

➡️Both B & D have more than one element,

➡️So, they are compound events.

$\Large\mathcal\green{FOLLOW \: ME}$

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Answer 2

Answer:

➡️Since either coin can turn up Head (H) or Tail (T), are the possible outcomes.

➡️But, now three coins are tossed once so the possible sample space contains,

➡️S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTH}

➡️Now,

➡️A: ‘three heads’

➡️A= (HHH)

➡️B: “two heads and one tail”

➡️B= (HHT, THH, HTH)

➡️C: ‘three tails’

➡️C= (TTT)

➡️D: a head shows on the first coin

➡️D= (HHH, HHT, HTH, HTT)

✴➡️(i) Mutually exclusive

➡️A ∩ B = (HHH) ∩ (HHT, THH, HTH)

➡️= φ✔

➡️Therefore, A and C are mutually exclusive.

➡️A ∩ C = (HHH) ∩ (TTT)

➡️= φ✔

➡️There, A and C are mutually exclusive.

➡️A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT)

➡️= (HHH)✔

➡️A ∩ D ≠ φ

➡️So they are not mutually exclusive

➡️B ∩ C = (HHT, HTH, THH) ∩ (TTT)

➡️= φ✔

➡️Since there is no common element in B & C, so they are mutually exclusive.

➡️B ∩ D = (HHT, THH, HTH) ∩ (HHH, HHT, HTH, HTT)

➡️= (HHT, HTH)✔

➡️B ∩ D ≠ φ

➡️Since there are common elements in B & D,

➡️So, they not mutually exclusive.

➡️C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT)

➡️= φ✔

➡️Since there is no common element in C & D,

➡️So they are not mutually exclusive.

✴➡️(ii) Simple event

➡️If an event has only one sample point of a sample space, it is called a simple (or elementary) event.

➡️A = (HHH)✔

➡️C = (TTT)✔

➡️Both A & C have only one element,

➡️so they are simple events.

✴➡️(iii) Compound events

➡️If an event has more than one sample point, it is called a Compound event

➡️B= (HHT, HTH, THH)✔

➡️D= (HHH, HHT, HTH, HTT)✔

➡️Both B & D have more than one element,

➡️So, they are compound events.

Step-by-step explanation:

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