## 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

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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2021-08-16T14:34:48+00:00
2021-08-16T14:34:48+00:00 2 Answers
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## Answers ( )

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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?━━━━━━━━━━━━━━━━━━━━━━━━━

➡️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.━━━━━━━━━━━━━━━━━━━━━━━━━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:## HOPE IT HELP YOU ✌✌