India plays two matches each with West Indies and Australia. In any match,the probabilities of India getting $0, 1,$ and $2$ points are $0.45, 0.05,$ and $0.50$ respectively. Assuming that the outcomes are independent,the probability of India getting at least $7$ points is:

  • A
    $0.8750$
  • B
    $0.0875$
  • C
    $0.0625$
  • D
    $0.0250$

Explore More

Similar Questions

Bismuth has a half-life of $5$ days. If a sample originally has a mass of $800 \text{ mg}$, then the mass remaining after $30$ days will be: (in $\text{ mg}$)

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome$\omega_1$$\omega_2$$\omega_3$$\omega_4$$\omega_5$$\omega_6$
$(a)$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{6}$

$A$ random variable $X$ has the following probability distribution:
$X = x$$1$$2$$3$$4$$5$$6$$7$$8$
$P(X = x)$$0.15$$0.23$$k$$0.10$$0.20$$0.08$$0.07$$0.05$

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$,then $P(E \cup F) = $

$A$ random variable $X$ has its range $\{-1, 0, 1\}$. If its mean is $0.2$ and $P(X=0)=0.2$,then $P(X=1)=$

$A$ fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$,$b=P(X \geq 3)$ and $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo