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

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33 votes

Best answer

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:

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

0

**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)

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