1 Answer

Espen Schulstad · Answer 1 · 2023-10-05T20:20:07+0000

Given:

One coin is tossed 3 times.

Required:

We need to find the probability of

a) 3 heads,

b) 3 tails and

c) tails then heads, then tails

Step-by-step explanation:

The sample space for tossing a coin 3 times

$S=\lbrace HHH,HTT.HTH.HHT.HTT,THH,TTT.TTH.HHT.TTT\rbrace$
n(S)=8

a)

Let A be the event of getting 3 heads.

$A=\lbrace HHH\rbrace$
n(A)=1

The probability of getting three heads is

$P(3\text{ heads\rparen=}(n(A))/(n(S))=(1)/(8)$

b)

Let B be the event of getting 3 tails.

$B=\lbrace TTT\rbrace$
n(B)=1

The probability of getting three tails is

$P(3\text{ tails\rparen=}(n(B))/(n(S))=(1)/(8)$

c)

Let C be the event of getting tails then heads, then tails

$C=\lbrace THT\rbrace$
n(C)=1

The probability of getting tails then heads, then tails is

$P(\text{ tails then heads, then tails\rparen=}(n(C))/(n(S))=(1)/(8)$

Final answer:

$P(3\text{ heads\rparen}=(1)/(8)$
$P(3\text{ tails\rparen}=(1)/(8)$
$P(\text{ tails then heads, then tails\rparen}=(1)/(8)$

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