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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. $\frac{15}{16}$

reshown

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

Probability of at least one head and one tail = 1- Probability of No (head and tail)

=  1 –  2 / (2*2*2*2)     {Numerator HHHH and TTTT and denominator all choices}

= 7/8

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 $= 1 - \dfrac{1}{16} - \dfrac{1}{16} = \dfrac{7}{8}.$

nice bro, we can also solve it using tree method

Using tree method it provides 14 favourable outcomes out of 16

So probability should be 14/16=7/8

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

Another simple approach:

and q= P(tails) = 1/2

Requirement:

Using binomial distribution,

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

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

$= \frac{7}{8}$

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

### 1 comment

1 head 3 tails =1/16 * 4C1 = 1/4

2 heads 2 tails =1/16 *4C2 = 3/8

3 heads 1 tail =1/16 *4C3 = 1/4