Question

A coin is tossed 5 times. what is the probability of getting head and and tail alternatively ​

Answer 1

Answer:

when a coin is tossed 5 times; number of possible outcomes = n(S) = 2^5 = 32. Now, more than 3 heads mean: (4 Heads + 1 Tail) and (5 Heads + 0 Tail).

Step-by-step explanation:

MATHS

A single coin is tossed 5 times. What is the probability of getting at least one head?

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ANSWER

Let us assume a coin is fair and 2-sided. A coin usually flips heads or tails. Then to flip heads or tails equals

2

1

. Consider flip a coin 5 times. Since a coin flip has two outcomes, then a coin flip 5 times has 2

5

=32 outcomes.

We may show the outcomes, e.g. use sample space S. Then

(1)S=(H,H,H,H,H),(H,H,H,H,T),…,(T,H,H,H,H),(T,T,T,T,T)

It must follow ∣S∣=32. A helpful way to see may be S=H,T

5

=H,Tx…xH,T. If you see the complement of at least 1 tails means 0 tails or 5 heads, then we must surely see 1 outcome, e.g. all heads. Then it must follow at least one tails has 31 outcomes,e.g.32–1=31.

To find or compute the probability, then #favorable/#possible in outcomes. It means 31/32, e.g. or P(E)=1–

32

1

=

32

31

.

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Answer 2

Let p be the probability of getting head in any toss and q = 1 - p, the probability of getting a tail. In 5 tosses, probability of getting head and tail alternately

$= P(THTHT) + P(HTHTH)$

$= q\times p\times q\times p\timesq - p\times q\timesp\times q\times p$

$= (P^2)(q^3) + (P^3)(q^2)$

$= (P^2)(q^2) (p+q)$

$= (P^2)(q^2).$

If $p = q= 1/2$, then the required probability $= 1/16$

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