Accountancy, asked by itsanjum24, 1 month ago

seven dice are thrown simultaneously. Find the probability :
(i) of getting one four times
(ii) of one atleast three times
(iii) of getting three on four dices atleast

Answers

Answered by aditipitale1
0

Answer:

The correct answer is 3)

Answered by soniatiwari214
0

Concept:

Probability can be defined as the ratio of favorable outcomes to the total possible outcomes.

Given:

Seven dice are thrown simultaneously.

Find:

The probability of getting one four times, of one at least three times, and of getting three on four dices at least.

Solution:

When Seven dice are thrown there are 6×6×6×6×6×6×6 = 6⁷, the number of possible outcomes.

(i) Getting one four times

Exactly four times 1 will come.

The favorable outcome, 1×1×1×1×5×5×5 = 125

Probability, P = 125/6⁷

(ii) Getting one at least three times

Getting 1 exactly three times + Getting 1 exactly four times + Getting 1 exactly five times + Getting 1 exactly six times + Getting 1 exactly seven times

The favorable outcome, 1×1×1×5×5×5×5 + 1×1×1×1×5×5×5 + 1×1×1×1×1×5×5 + 1×1×1×1×1×1×5 + 1×1×1×1×1×1×1 = 625 + 125 + 25 + 5 + 1 = 781

Probability, P = 781/6⁷

(ii) Getting three at least four times

Getting 3 exactly four times + Getting 3 exactly five times + Getting 3 exactly six times + Getting 3 exactly seven times

The favorable outcome, 1×1×1×1×5×5×5 + 1×1×1×1×1×5×5 + 1×1×1×1×1×1×5 + 1×1×1×1×1×1×1 = 125 + 25 + 5 + 1 = 156

Probability, P = 156/6⁷

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