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)

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• 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:
\[
P = \frac{14}{16} = \frac{7}{8}.
\]
 

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Beutiful

Answer 1

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

Comments

nice bro, we can also solve it using tree method

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Using tree method it provides 14 favourable outcomes out of 16 

So probability should be 14/16=7/8

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absolute blessing to the human race

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someone please explain what is tree method?

Answer 2

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

Comments

Simply Clever.

Answer 3

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

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

$= \frac{7}{8}$

Comments

very good answer by using the concept

Answer 4

Let $X$ be a binomial random variable representing the number of Heads in $n = 4$ independent fair coin tosses, where the probability of success (getting a head) is $p = \frac{1}{2}$ and failure is $q = \frac{1}{2}$.

The condition "at least one head and at least one tail" means the number of heads must satisfy $1 \le X \le 3$.

Using the complement rule:

$$P(1 \le X \le 3) = 1 - [P(X = 0) + P(X = 4)]$$

Using the Binomial PMF $P(X = k) = \binom{n}{k} p^k q^{n-k}$:

• For 0 heads (All tails): $P(X = 0) = \binom{4}{0} \left(\frac{1}{2}\right)^4 = \frac{1}{16}$
• For 4 heads (All heads): $P(X = 4) = \binom{4}{4} \left(\frac{1}{2}\right)^4 = \frac{1}{16}$

 

Substituting these back into the complement equation:

$$P(1 \le X \le 3) = 1 - \left( \frac{1}{16} + \frac{1}{16} \right) = 1 - \frac{2}{16} = \frac{14}{16} = \frac{7}{8}$$

Hence, option C is the correct answer.

Answer 5

Ans

Answer 6

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

Comments

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.