Question

A coin is tossed thrice find the probability of getting exactly two heads or atleast one tail or consecutive two head

Answer 1

Answer:

1)1/3

2)1/6

3)1/2

Step-by-step explanation:

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Answer 2

Probability

Normally, We will get a head and a tail while tossing a coin. Here to find the probability of a coin tossed thrice, which means the coin is tossed three times. So, it's probability would be,

(Refer the attachment for the method)

Sample Space = {HHH, HHT, HHT, HTT, THH, THT, TTH, TTT}

n(S) = 8

We know that,

1) A - is the event of getting exactly 2 heads

$⟹$A = {HHT, HTH, THH}

$⟹$n(A) = 3

$⟹$P(A) = $\sf{\dfrac{n(A)}{n(S)}}$

$⟹$P(A) = $\sf{\dfrac{3}{8}}$

2) B - is the event of getting at least one tail

$⟹$ B = {TTT, TTH, HTT, THT, HHT, HTH, THH}

$⟹$n(B) = 7

$⟹$P(B) = $\sf{\dfrac{n(B)}{n(S)}}$

$⟹$P(B) = $\sf{\dfrac{7}{8}}$

3) C - Event of getting consecutively 2 heads

$⟹$C = {HHT, THH, HHH}

$⟹$n(C) = 3

$⟹$P(C) = $\sf{\dfrac{n(C)}{n(S)}}$

$⟹$P(C) = $\sf{\dfrac{3}{8}}$

4) A∩B = {HHT, HTH, THH}

$⟹$n(A∩B) = 3

$⟹$P(A∩B) = $\sf{\dfrac{n(A∩B)}{n(S)}}$

$⟹$P(A∩B) = $\sf{\dfrac{3}{8}}$

5) B∩C = {HHT, THH}

$⟹$n(B∩C) = 2

$⟹$P(B∩C) = $\sf{\dfrac{n(B∩C)}{n(S)}}$

$⟹$P(B∩C) = $\sf{\dfrac{2}{8}}$

6) A∩C = {HHT, THH}

$⟹$n(A∩C) = 2

$⟹$P(A∩C) = $\sf{\dfrac{n(A∩C)}{n(S)}}$

$⟹$P(A∩C) = $\sf{\dfrac{2}{8}}$

7) A∩B∩C = {HHT, THH}

$⟹$n(A∩B∩C) = 2

$⟹$P(A∩B∩C) = $\sf{\dfrac{n(A∩B∩C)}{n(S)}}$

$⟹$P(A∩B∩C) = $\sf{\dfrac{2}{8}}$

Solving using Formula :

$⟹$P(A∩B∩C) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(A∩C) + P(A∩B∩C)

$⟹$P(A∩B∩C) = $\sf{\dfrac{3}{8}}\:+\:$$\sf{\dfrac{7}{8}}\:+\:$$\sf{\dfrac{3}{8}}\:-\:$$\sf{\dfrac{3}{8}}\:-\:$$\sf{\dfrac{2}{8}}\:-\:$$\sf{\dfrac{2}{8}}\:+\:$$\sf{\dfrac{2}{8}}$

$⟹$P(A∩B∩C) = $\sf{\dfrac{3}{8}}\:+\:$$\sf{\dfrac{7}{8}}\:+\:$$\sf{\dfrac{{\cancel{3}}}{{\cancel{8}}}}\:-\:$$\sf{\dfrac{{\cancel{3}}}{{\cancel{8}}}}\:-\:$$\sf{\dfrac{2}{8}}\:-\:$$\sf{\dfrac{{\cancel{2}}}{{\cancel{8}}}}\:+\:$$\sf{\dfrac{{\cancel{2}}}{{\cancel{8}}}}$

$⟹$P(A∩B∩C) = $\sf{\dfrac{3}{8}}\:+\:$$\sf{\dfrac{7}{8}}\:+\:$$\sf{\dfrac{2}{8}}$

$⟹$P(A∩B∩C) = $\sf{\dfrac{10\:-\:2}{8}}$

$⟹$P(A∩B∩C) = $\sf{\dfrac{8}{8}}$

$⟹$P(A∩B∩C) = $\sf{1}$

Hence, the probability of getting exactly two heads or at least one tail or two consecutive heads is 1.