MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Probability

If P E 1 =0...

Question

If $P (E_{1}) = 0.2, P (E_{2}) = 0.4$ and $P (E_{3}) = 0.6$ and $E_{1}, E_{2}$ and $E_{3}$ are independent events, then the probability that at least one of these events $E_{1}, E_{2}$ and $E_{3}$ occurs is

A

1.2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

0.048

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C

0.808

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

D

0.952

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $0.808$
$P ($ at least one event $E_{1}, E_{2}, E_{3}$ occurs $)$
$= P (E_{1} \cup E_{2} \cup E_{3}) = P (E_{1}) + P (E_{2}) + P (E_{3}) - P (E_{1} \cap E_{2}) - P (E_{2} \cap E_{3}) - P (E_{3} \cap E_{1}) + P (E_{1} \cap E_{2} \cap E_{3}) = P (E_{1}) + P (E_{2}) + P (E_{3}) - P (E_{1}) P (E_{2}) - P (E_{2}) P (E_{3}) - P (E_{3}) P (E_{1}) + P (E_{1}) P (E_{2}) P (E_{3}) = 0.2 + 0.4 + 0.6 - (0.2) (0.4) - (0.4) (0.6) - (0.6) (0.2) + (0.2) (0.4) (0.6) = 0.808$

Suggest Corrections

thumbs-up

1

Similar questions

Q. Of the three independent events

E_{1}

E_{2}

E_{3}

, the probability that only

E_{1}

α

E_{2}

β

E_{3}

γ

. Let the probability

p

that none of events

E_{1}

E_{2}

E_{3}

occurs satisfy the equations

(α - 2 β) p = α β

(β - 3 γ) p = 2 β γ

. All the given probabilities are assumed to lie in the interval

(0, 1)

\frac{Probability of occurrence of E_{1}}{Probability of occurrence of E_{3}} =

Events $E_{1}, E_{2}, E_{3}$ are the possible events of an experiment and their probabilities are recorded. Mark the possible correct answer

E_{1}, E_{2}, E_{3}

are the possible events of an experiment and their probabilities are recorded. Mark the possible correct answer

The probability that at least one of the events $E_{1}$ and $E_{2}$ occurs is 0.6 If the probability of the simultaneous occurrence of $E_{1}$ and $E_{2}$ is 0.2, find $P ({¯ E}_{1}) + P ({¯ E}_{2}) .$

E^{c}

denote the complement of an event

E

E_{1}, E_{2}

E_{3}

be any pairwise independent events with

P (E_{1}) > 0

P (E_{1} \cap E_{2} \cap E_{3}) = 0

P (\frac{E_{2}^{c} \cap E_{3}^{c}}{E_{1}})

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Probability

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon