edited by
1,722 views
1 votes
1 votes

A person on a trip has a choice between a private car and public transport. The probability of using a private car is $0.45.$ While using public transport, the further choice available are bus and metro. Out of which the probability of commuting by a bus is $0.55.$ In such a situation, the probability (rounded up to two decimals) of using a car, bus and metro respectively would be

  1. $0.45, 0.30$ and $0.25$
  2. $0.45, 0.25$ and $0.30$
  3. $0.45, 0.55$ and $0$
  4. $0.45, 0.35$ and $0.20$
edited by

2 Answers

0 votes
0 votes
They have given that probability of choosing car which is 0.45.

And hence probability of choosing public transport=1 -0.45=0.55 which is going to be distributed among Bus and metro.

They have given that out of 0.55, probability of choosing bus = 0.55 * 0.55 = 0.3025=0.30(Approx)

So , probability of choosing metro will be=0.55 * 0.45 =0.2475= 0.25 (Approx)

I think Option A will be Correct.
Answer:

Related questions

3 votes
3 votes
1 answer
1
sh!va asked Feb 22, 2017
418 views
Evaluate $\int_0^1 \int_0^{\sqrt{1+x^2}} \frac{d x \cdot d y}{\left(1+x^2+y^2\right)}$$\frac{\pi}{2}[\log (1+\sqrt{2})]$$\frac{\pi}{4}[\log (1+\sqrt{2})]$$\frac{\pi}{2}[\...
1 votes
1 votes
3 answers
2
sh!va asked Feb 22, 2017
424 views
The value of $\lim _{x \rightarrow 8}\left(\frac{x^{1 / 3}-2}{x-8}\right)$ is$1/4$$1/8$$1/12$$1/16$
1 votes
1 votes
1 answer
3
sh!va asked Feb 22, 2017
615 views
in given combinational logic $X=?$$X = AB'C' +A'BC'+ A'B'C+ ABC$$X = A'BC +A'BC'+ AB'C+ A'B'C'$$X = AB + BC + AC$$X= A'B'+B'C'+A'C'$
1 votes
1 votes
0 answers
4
sh!va asked Feb 22, 2017
670 views
$000, 001, 010, 011, 100\; \&$ repeats$100, 011, 010, 001, 000\; \&$ repeat$010, 011, 100, 000, 001\; \&$ repeats$101, 110, 111, 000, 001, 010, 011, 100 \;\&$ repeats