Q.An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X? Is X a random variable? OR Bag I contains 3 red and 4 black balls and Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from Bag II.

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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