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The probability of three consecutive heads in four tosses of a fair coin is

1. $\left(\dfrac{1}{4}\right)$
2. $\left(\dfrac{1}{8}\right)$
3. $\left(\dfrac{1}{16}\right)$
4. $\left(\dfrac{3}{16}\right)$
5. None of the above

Best one praggy sir.

Just copying yours but some little modification.

H and H and ( 1 - No_Head)

i.e H and H and ( 1 - (T and T ))

1/2* 1/2 * ( 1- (1/2*1/2))

3/16
What is 1st one is Tail and then 3 consecutive Head , then you can't get 3/16.

Let the $4$ tosses be named $P,Q,R$ and $S$

To have $3$ consecutive heads:

• $Q$, $R$ must be both heads.
• At least one of $\left \{ P, S \right \}$ must be a head.

Thus, the probability of getting 3 consecutive heads is given by:

$P = P_q \times P_r \times \underbrace{\left ( P_p + P_s - P_p P_s \right )}_{\text{atleast one}}$

$\quad= \dfrac 12 \times \dfrac 12 \times \left ( \dfrac 12 + \dfrac 12 - \dfrac 12 \cdot \dfrac 12 \right )$

$P = \frac{3}{16}.$

Hence, option D is the correct answer.

Another way of looking at it is:

$P = P_{HHHT} + P_{THHH} + P_{HHHH} = \dfrac{1}{16} + \dfrac{1}{16} + \dfrac{1}{16} = \dfrac{3}{16}$

sir, why cant i do it like below

suppose n =3 (no of times exp repeated)

P(HHHT)   = 1/2^4

P(3 Heads) = 4C3* 1/2^4 = 1/4

How to screw up?

Let A = Probability of getting first 3 tossings as head

Let B = Probability of getting last 3 tossings as head

P(A) + P(B) = (1/2)$^{3}$ + (1/2)$^{3}$ = (1/4) ----> {This is WRONG}

This is not the correct answer because we have counted Probability of getting all 4 tossings as head twice.

So, P(A$\cup$B) = P(A) + P(B) - P(A$\cap$B) = (1/2)$^{3}$ + (1/2)$^{3}$ - (1/2)$^{4}$ = (3/16)

How it is -(Pp*Ps) in at least one part ?

At least means ( Pp or Ps or Both)

Therefore It should be +(Pp*Ps).

But why it is wrong here ?

Number of outcomes which having 3 consecutive Heads = {HHHT , THHH, HHHH} = 3

Total no of possible outcomes with 4 tosses = 24 =16

P(having 3 consecutive Heads) = 3 /16

## The correct answer is (d) 3/16

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same explanation as best answer !!!!

We can also do this question with the help of the tree method,

simply try to write down
$HHHT$, $THHH$
AND THE LAST ONE, POSSIBLY YOU MAY MISS IT
$HHHH$

SAMPLE SPACE : 16
POSSIBLE OUTCOMES : 3
$3/16$ ANSWER

$EXACTLY$ $3$ $CONSECUTIVE$ $HEADS$
$2/16$

I.E. $1/8$

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