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EXERCISE 8.1 · Q5

Q.The probability that a bulb produced by a factory will fuse after 150 days of use is 0.050.05. Find the probability that out of 5 such bulbs

(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use.
Maharashtra MsbshseTextbookSubjectiveImportance★★★★★
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With n=5n=5 bulbs and probability of fusing p=0.05p=0.05 (so q=0.95q=0.95), X∼B(5,0.05)X\sim B(5,0.05) counts how many of the 5 bulbs fuse.

  1. None fuse: P(X=0)=(0.95)5≈0.7738P(X=0)=(0.95)^5\approx0.7738.
  2. Not more than one: P(X≤1)=P(0)+P(1)=(0.95)5+5(0.05)(0.95)4≈0.7738+0.2036=0.9774P(X\le1)=P(0)+P(1)=(0.95)^5+5(0.05)(0.95)^4\approx0.7738+0.2036=0.9774.
  3. More than one: P(X>1)=1−P(X≤1)≈1−0.9774=0.0226P(X>1)=1-P(X\le1)\approx1-0.9774=0.0226. …

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