GATE CSE 2002 | Question: 2.16

Question

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

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{7}{8}$
• D. $\frac{15}{16}$

Comments

saying atleast 1H and 1T= All head and All tails are not allowed (HHHH and TTTT are not allowed)

---

 

• Only two outcomes, \(\text{HHHH}\) and \(\text{TTTT}\), do not contain both H and T i.e. combination of H and T.  
   
• Out of \(16\) total outcomes, the remaining \(14\) do.  

Therefore, the probability is:
\[
P = \frac{14}{16} = \frac{7}{8}.
\]
 

---

Beutiful

Answer 1

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

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

---

Simplest Approach!