GO Classes DPP | GATE DA | Probability | Conditional Probability

Answer 1

When $3$ fair coins are flipped, the total sample space is

$S=\{HHH,\ HHT,\ HTH,\ THH,\ HTT,\ THT,\ TTH,\ TTT\}.$

We are given that $\textit{at least one coin shows a head and at least one coin shows a tail}.$

So the outcomes $HHH$ and $TTT$ are not possible.

Hence, the reduced sample space is

$S'=\{HHT,\ HTH,\ THH,\ HTT,\ THT,\ TTH\}.$

Thus, the number of possible outcomes is

$n(S')=6.$

Now we need the probability that $\textit{exactly two of the coins show heads}.$

The outcomes with exactly two heads are

$\{HHT,\ HTH,\ THH\}.$

So the number of favorable outcomes is $3.$

$\therefore P(\text{exactly two heads} \mid \text{at least one head and at least one tail})=\frac{3}{6}=\frac{1}{2}.$

Answer: $\boxed{\dfrac{1}{2}}$

Answer 2

Given, 

3 fair coins are flipped with at least 1H and 1T.

So, Sample Space S = {HHT, HTH, HTT, THH, THT, TTH} (total of 6 outcomes) since HHH and TTT violates the condition of atleast 1H and 1T.

Now, 

Conditions satisfying atleast 2H are: HHT, HTH, THH (3 outcomes).

So, 3/6 = 1/2

Option C is the correct answer.

Answer 3

Answer: C

Answer 4

Let A be the event of exactly two coins showing heads and 

B be the event of  atleast one coin showing head and atleast one coin showing tail
 

Now, we have to find out $\textbf{P(A | B)}$

$P(A | B )  = \frac{P(A,B)}{P(B)}$

$P(A,B) = \frac{3}{2^3} = $ P(2 heads, 1 tail)

$P(B) = $ P(atleast one head, atleast one tail) = 1 - P(0 heads) - P(0 tails) = $1 - \frac{1}{8} - \frac{1}{8} = \frac{6}{8}$ 

Now, $P(A | B ) = \frac{\frac{3}{2^3}}{\frac{6}{8}} = \frac{3}{6} = \frac{1}{2}$

Hence, Answer is Option C