Q.Two fair coins are tossed simultaneously. Let $X$ denote the number of heads obtained. Write the probability distribution of $X$ and find $E(X)$.

All probabilities in a distribution must sum to 1.

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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; …

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 …

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

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

$\sum p(x) = 1$ …

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 $1$:

$k+2k+3k=1$

$6k=1$ …

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

The probability distribution is

Since total probability is 1: …

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:

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

$0.6+3k=1$ …

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):

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

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

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

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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 $1$:

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

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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

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

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