Q.Three fair coins are tossed simultaneously. Let $X$ denote the number of tails obtained. Find the probability distribution of $X$.

All probabilities in a distribution must sum to 1.

$k+2k+3k+4k+5k = 15k = 1 \Rightarrow k = \dfrac{1}{15}$

The $r$th head lands on toss $n$ means: exactly $r-1$ heads in the first $n-1$ tosses, and the $n$th toss is a head.

This is a Bernoulli-trials counting argument, an NCERT/CBSE Class 12 probability staple (here with success probability $p$, $q=1-p$).

For the $r$th head to appear precisely on the $n$th toss:

• among the first $n-1$ tosses there are exactly $r-1$ heads (and $(n-1)-(r-1)=n-r$ tails), which happens in $^{n-1}C_{r-1}\,p^{r-1}q^{n-r}$ ways/probability;
• the $n$th toss is a head, probability $p$.

Multiplying the independent parts:

$^{n-1}C_{r-1}\,p^{r-1}q^{n-r}\cdot p = {}^{n-1}C_{r-1}\,p^{r}q^{n-r}.$

Find $k$ from $\sum P=1$, then add the probabilities for $X=2,3,4$.

Using $\sum P(X=x_i)=1$ to complete a probability distribution is a CBSE/NCERT Class 12 probability skill.

Since all probabilities sum to $1$:

$k+2k+3k+4k+5k = 15k = 1 \Rightarrow k = \frac{1}{15}.$

Then

$P(X\ge2) = P(2)+P(3)+P(4) = 3k+4k+5k = 12k = \frac{12}{15} = \frac{4}{5}.$

Use the fact that the total probability of a probability distribution equals 1.

For $p(x)$ to be a valid probability distribution:

$\sum p(x) = 1$

$0.3 + k + 0.2 + 0.1 = 1$

$0.6 + k = 1 \implies k = 0.4$

Use that the total probability over all values of $X$ equals $1$.

The distribution assigns $P(X=0)=k,\ P(X=1)=2k,\ P(X=2)=3k$, and $0$ elsewhere.

Since the probabilities of all possible outcomes must add to $1$:

$k+2k+3k=1$

$6k=1$

$k=\frac16.$

(Each of $k,2k,3k$ then lies in $[0,1]$, so this is a valid distribution.)

First find K using $\sum P(X)=1$, then add $P(0)+P(1)$.

The probability distribution is

Since total probability is 1:

$K+2K+3K+4K = 1 \ \Rightarrow\ 10K=1 \ \Rightarrow\ K=0.1$

$P(X<2)$ means $X=0$ or $X=1$:

$P(X<2) = P(0)+P(1) = K+2K = 3K = 3(0.1) = 0.3$

Since total probability must equal 1, solving $0.6+3k=1$ gives $k=2/15$.

For a probability distribution, all probabilities must sum to 1:

$0.1+k+0.3+2k+0.2=1$

$0.6+3k=1$

$3k=0.4$

$k=\dfrac{0.4}{3}=\dfrac{2}{15}$

This is a binomial probability — use $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$ with $n=3$, $p=\frac12$, $k=1$.

An unbiased coin is tossed $3$ times, so $n=3$ and $p=P(\text{head})=\dfrac12$ for each toss.

Probability of exactly $1$ head (binomial distribution formula):

$P(X=1) = \binom{3}{1}\left(\dfrac12\right)^1\left(\dfrac12\right)^{2}$

$\binom{3}{1}=3$

$P(X=1) = 3\times\dfrac12\times\dfrac14 = \dfrac{3}{8}$

(This also makes sense by direct counting: out of $2^3=8$ equally likely outcomes, exactly $3$ have precisely one head — HTT, THT, TTH.)

Setting the sum of probabilities to $1$ gives a quadratic in $k$; the valid probability root is $k=\tfrac13$.

For a probability distribution, all probabilities must sum to $1$:

$k+k^2+2k^2+k=1\ \Rightarrow\ 3k^2+2k-1=0.$

Solve: $k=\dfrac{-2\pm\sqrt{4+12}}{6}=\dfrac{-2\pm4}{6}$, giving $k=\dfrac13$ or $k=-1$.

Since a probability cannot be negative, $k=-1$ is rejected.

Use the binomial probability formula with $n=10$, $p=1/2$; note $^{10}C_6=\,^{10}C_4$.

For a fair coin tossed 10 times, $X\sim B(10,1/2)$, so

$P(X=6)=\binom{10}{6}\left(\dfrac12\right)^{10}$

Since $\binom{10}{6}=\binom{10}{10-6}=\binom{10}{4}$ (by the symmetry property of binomial coefficients), this equals $^{10}C_4\cdot\dfrac{1}{2^{10}}$.