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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. $\frac{15}{16}$
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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

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25 votes
 
Best answer

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

2 Comments

nice bro, we can also solve it using tree method
0

Using tree method it provides 14 favourable outcomes out of 16 

So probability should be 14/16=7/8

0
27 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

13 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 vote
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

So, adding gives 7/8. 

 

11
–1 vote
probability of getting head=p=1/2;

by the formula p(x)=nCx*P^xq^(n-x)

p(atleast one tail)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16

p(atleast one head)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16

so

P(head&tail)=15/16*15/16=7/8

2 Comments

number of probabilities of 4 coins = 2⁴ = 16

We have a formula that
         P(x≥1)+P(y≥1)

               = 1-(P(x<1)+P(y<1))

               = 1-(P(x=0)+P(y=0))

               = 1-((1/16)+(1/16))
               = 1-(2/16)
               = 7/8
2

Simplest Approach!

2
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