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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. $\frac{15}{16}$
asked in Probability by Veteran (59.9k points) | 2.3k views
+6

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

5 Answers

+19 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 $= 1 - \dfrac{1}{16} - \dfrac{1}{16} = \dfrac{7}{8}.$

answered by Loyal (8.8k points)
edited by
0
nice bro, we can also solve it using tree method
+13 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

answered by Boss (24.9k points)
0 votes
If you didn't understand the methods above,

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

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

$= \frac{7}{8}$
answered by Active (1.5k points)
–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
answered by Boss (14.5k points)
+9

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. 

 

–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
answered by (115 points)
+2
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
0

Simplest Approach!

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