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14 votes
14 votes

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
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4 Answers

Best answer
33 votes
33 votes

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:

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

edited by
3 votes
3 votes

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

0 votes
0 votes
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

HAD IT BEEN

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

I.E. $1/8$

WOULD BE THE ANSWER.
Answer:

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