Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are 2

Q: Wait, why do we set the discriminant to zero? Can't it be greater than zero for real roots?

Great question! The problem specifically asks for 'real and equal' roots. If the discriminant were greater than zero, the roots would be real but distinct. For them to be equal, the parabola must just touch the x-axis, which happens only when $D=0$.

Q: Why did we reject $\beta = 1/4$ when $\alpha = 1$?

Remember, $\beta$ is defined as the number of heads in two tosses. You cannot have a quarter of a head! Since $\beta$ must be an integer from the set $\{0, 1, 2\}$, any value outside this set is physically impossible in this context.

Q: Why do we multiply the probabilities for the pairs?

Because the tosses of Coin 1 and Coin 2 are independent events. The outcome of Coin 1 does not influence Coin 2. When we want two independent events to happen together (an 'AND' condition), we multiply their individual probabilities.

Question

Let $C_1$ and $C_2$ be two biased coins such that the probabilities of getting head in a single toss are $\frac{2}{3}$ and $\frac{1}{3}$, respectively. Suppose $\alpha$ is the number of heads that appear when $C_1$ is tossed twice, independently, and suppose $\beta$ is the number of heads that appear when $C_2$ is tossed twice, independently. Then probability that the roots of the quadratic polynomial $x^2 - \alpha x + eta$ are real and equal, is

Options:

(A) $\frac{40}{81}$

(B) $\frac{20}{81}$

(C) $\frac{1}{2}$

(D) $\frac{1}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a quadratic equation, $x^2 - \alpha x + \beta = 0$. The problem asks us to find the probability that this equation has real and equal roots.

In the world of algebra, this is a call to action for the discriminant. We know that for any quadratic $ax^2 + bx + c = 0$, the nature of the roots is governed by $D = b^2 - 4ac$.

For the roots to be real and equal, the parabola must kiss the x-axis at exactly one point, which means $D$ must be zero. Substituting our coefficients, where $a=1$, $b=-\alpha$, and $c=\beta$, we get:

$$(-\alpha)^2 - 4(1)(\beta) = 0$$

This simplifies to the elegant constraint:

$$\alpha^2 = 4\beta$$

### The Discrete Search

Now, let's step into the shoes of a probability theorist. We are tossing two biased coins. $\alpha$ is the number of heads from two tosses of Coin 1, and $\beta$ is the number of heads from two tosses of Coin 2.

Since each coin is tossed twice, the possible values for $\alpha$ and $\beta$ are restricted to the set $\{0, 1, 2\}$. We need to find pairs $(\alpha, \beta)$ that satisfy $\alpha^2 = 4\beta$.

If $\alpha = 0$, then $0^2 = 4\beta$, forcing $\beta = 0$. That is our first valid pair: $(0, 0)$.

If $\alpha = 1$, then $1^2 = 4\beta$, which gives $\beta = 1/4$. Since $\beta$ must be an integer, we must reject this.

If $\alpha = 2$, then $2^2 = 4\beta$, which gives $4 = 4\beta$, so $\beta = 1$. That is our second valid pair: $(2, 1)$.

### The Binomial Engine

We have our targets: $(0, 0)$ and $(2, 1)$. Now, we calculate the probability of these events using the binomial distribution:

$$P(X=r) = \binom{n}{r} p^r q^{n-r}$$

For Coin 1, $P(H) = 2/3$ and $P(T) = 1/3$. Thus:

$$P(\alpha=0) = \binom{2}{0} (2/3)^0 (1/3)^2 = 1/9$$

$$P(\alpha=2) = \binom{2}{2} (2/3)^2 (1/3)^0 = 4/9$$

For Coin 2, $P(H) = 1/3$ and $P(T) = 2/3$. Thus:

$$P(\beta=0) = \binom{2}{0} (1/3)^0 (2/3)^2 = 4/9$$

$$P(\beta=1) = \binom{2}{1} (1/3)^1 (2/3)^1 = 4/9$$

### The Final Convergence

Since the coin tosses are independent, the probability of a pair occurring is the product of the individual probabilities. The total probability is $P(\alpha=0)P(\beta=0) + P(\alpha=2)P(\beta=1)$.

Substituting our values, we get:

$$(1/9 	imes 4/9) + (4/9 	imes 4/9) = 4/81 + 16/81 = 20/81$$

The final probability is **$20/81$**.

Select Answer:

Visualized Solution

$T h e Q u a d r a t i c E q u a t i o n$

$C o n d i t i o n f or R e a l an d E q u a l R oo t s$

$A ppl y in g t h eD i scr iminan tC o n d i t i o n$

$D e f inin g t h e S am pl e S p a ce$

$F in d in g V a l i d P ai r s : C a se α = 0$

$F in d in g V a l i d P ai r s : C a ses α = 1, 2$

$S u mma r y o f V a l i d P ai r s$

$P r o babi l i t y D i s t r ib u t i o n f or C o in 1$

$P r o babi l i t y D i s t r ib u t i o n f or C o in 2$

$C o mbinin g I n d e p e n d e n tE v e n t s$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Binomial Distribution

Analyzing the Setup

The Discrete Search

The Binomial Engine

The Final Convergence

Similar Questions

A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is :

One hundred identical coins, each with probability, $p$ , of showing up heads are tossed once. If $0 < p < 1$ and the probability of heads showing on $50$ coins is equal to that of heads showing on $51$ coins, then the value of $p$ is

Let a die be rolled $n$ times. Let the probability of getting odd numbers seven times be equal to the probability of getting odd numbers nine times. If the probability of getting even numbers twice is $\frac{k}{2 ^{15}}$ , then $k$ is equal to

If an unbiased die, marked with $- 2, - 1, 0, 1, 2, 3$ on its faces, is thrown five times, then the probability that the product of the outcomes is positive, is

Let $X$ be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$ . Then $54 P (X \leq 2)$ is equal to

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is :

If a random variable $X$ follows the Binomial distribution $B (33, p)$ such that $3 P (X = 0) = P (X = 1)$ , then the value of $\frac{P ( X = 15 )}{P ( X = 18 )} - \frac{P ( X = 16 )}{P ( X = 17 )}$ is equal to

The random variable $X$ follows binomial distribution $B (n, p)$ , for which the difference of the mean and the variance is 1. If $2 P (X = 2) = 3 P (X = 1)$ , then $n^{2} P (X > 1)$ is equal to

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is :

An ordinary dice is rolled for a certain number of times. If the probability of getting an odd number 2 times is equal to the probability of getting an even number 3 times, then the probability of getting an odd number for odd number of times is :

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TheQuadraticEquation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ConditionforRealandEqualRoots

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ApplyingtheDiscriminantCondition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } DefiningtheSampleSpace

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingValidPairs:Caseα=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FindingValidPairs:Casesα=1,2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } SummaryofValidPairs

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ProbabilityDistributionforCoin1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ProbabilityDistributionforCoin2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CombiningIndependentEvents

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } FinalCalculation