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22 votes
22 votes

Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is

  1. $\frac{1}{16}$
  2. $\frac{1}{8}$
  3. $\frac{7}{8}$
  4. $\frac{15}{16}$

5 Answers

Best answer
36 votes
36 votes

Answer - C

probability of getting all heads =$\dfrac{1}{16}$

probability of getting all tails =$\dfrac{1}{16}$

probability of getting at least one head and one tail $= 1 - \dfrac{1}{16} - \dfrac{1}{16} = \dfrac{7}{8}.$

edited by
35 votes
35 votes

Total outcomes - 24  (Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)
Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.

So, probability, favourable events/total outcome

14/16 = 7/8

16 votes
16 votes

Another simple approach:

Let p= P(heads) = 1/2

and q= P(tails) = 1/2

Requirement:

    1 Heads 3 Tails

or 2 Heads 2 Tails

or 3 Heads 1 Tails

Using binomial distribution,

Required probability = $_{}^{4}\textrm{C}_{1} p^{1} q^{3} + {}^{4}\textrm{C}_{2} p^{2} q^{2} + {}^{4}\textrm{C}_{3} p^{3} q^{1}$

= $_{}^{4}\textrm{C}_{1} (1/2)^{1} (1/2)^{3} + {}^{4}\textrm{C}_{2} (1/2)^{2} (1/2)^{2} + {}^{4}\textrm{C}_{3} (1/2)^{3} (1/2)^{1}$

$= \frac{7}{8}$

edited by
–1 votes
–1 votes
1 head 3 tails =1/16 2 heads 2 tails =1/16 3 heads 1 tail =1/16 adding we get 3/16 where i am wrong anyone plz explain
Answer:

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