The probability that a bomb hits a target is 0.4. three bombs can destroy a bridge. if in all 5 bombs are thrown find the probability that the bridge will be destroyed
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Probability that a bomb hits a target = 0.4 = 2/5
Probability that 3 bombs can destroy a bridge = 1.0
Or, to destroy a bridge at least three bombs must hit the bridge. So 3 or 4 or 5 bombs must hit the bridge.
5 bombs are thrown. Each one can be a Hit or Miss. It needs 3 H & 2 M, 4 H & 1 M or all 5 H. The number of combinations (permutations don't matter) are respectively:
5C3 , 5C4, 5C5
= 10, 5, 1.
Probability of a Hit = P(H) = 0.4
Probability of a Miss = P(M) = 0.6
So probability of 3 Hits & 2 Misses
= 10 * [P(H)]³ * [P(M)]² = 10 * 0.064 * 0.36 = 0.2304
Probability of 4 Hits & 1 Miss:
= 5 * 0.4⁴ * 0.6 = 0.0768
Probability of 5 Hits & 0 Misses:
= 1 * 0.4⁵ = 0.01024
Total: 0.31744
Probability that 3 bombs can destroy a bridge = 1.0
Or, to destroy a bridge at least three bombs must hit the bridge. So 3 or 4 or 5 bombs must hit the bridge.
5 bombs are thrown. Each one can be a Hit or Miss. It needs 3 H & 2 M, 4 H & 1 M or all 5 H. The number of combinations (permutations don't matter) are respectively:
5C3 , 5C4, 5C5
= 10, 5, 1.
Probability of a Hit = P(H) = 0.4
Probability of a Miss = P(M) = 0.6
So probability of 3 Hits & 2 Misses
= 10 * [P(H)]³ * [P(M)]² = 10 * 0.064 * 0.36 = 0.2304
Probability of 4 Hits & 1 Miss:
= 5 * 0.4⁴ * 0.6 = 0.0768
Probability of 5 Hits & 0 Misses:
= 1 * 0.4⁵ = 0.01024
Total: 0.31744
kvnmurty:
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hey there,
9down voteacceptedYour first calculation finds the probability that the person hits the target 44 times in a row. That is very different (and much smaller) than the probability that the person hits at least once.Let's do the problem in another way, much too long, but it will tell us what is going on. What is the probability of hitting at least once? The event "at least one hit" could happen in several ways: (i) exactly 11 hit; (ii) exactly 22 hits; (iii) exactly 33 hits; (iv) exactly 44 hits.(i) The probability of exactly one hit is (41)(1/4)(3/4)3(41)(1/4)(3/4)3 . This is because the hit could happen in any one of 44 (that is, (41)(41) ) places. Write H for hit and M for miss. The probability of the pattern HMMM is (1/4)(3/4)(3/4)(3/4)(1/4)(3/4)(3/4)(3/4) . Similarly, the probability of MHMM is (3/4)(1/4)(3/4)(3/4)(3/4)(1/4)(3/4)(3/4) . You will notice this probability is the same as the probability of HMMM. We get the same probability for MMHM and for MMMH, for our total of (41)(1/4)(3/4)3(41)(1/4)(3/4)3 .(ii) Similarly, the probability of exactly 22 hits is (42)(1/4)2(3/4)2(42)(1/4)2(3/4)2 . (iii) The probability of 33 hits is (43)(1/4)3(3/4)(43)(1/4)3(3/4) .(iv) The probability of 44 hits is (44)(1/4)4(44)(1/4)4 . This is the (1/4)4(1/4)4 that you calculated.Add up. We get the required answer. Hope this helps!
9down voteacceptedYour first calculation finds the probability that the person hits the target 44 times in a row. That is very different (and much smaller) than the probability that the person hits at least once.Let's do the problem in another way, much too long, but it will tell us what is going on. What is the probability of hitting at least once? The event "at least one hit" could happen in several ways: (i) exactly 11 hit; (ii) exactly 22 hits; (iii) exactly 33 hits; (iv) exactly 44 hits.(i) The probability of exactly one hit is (41)(1/4)(3/4)3(41)(1/4)(3/4)3 . This is because the hit could happen in any one of 44 (that is, (41)(41) ) places. Write H for hit and M for miss. The probability of the pattern HMMM is (1/4)(3/4)(3/4)(3/4)(1/4)(3/4)(3/4)(3/4) . Similarly, the probability of MHMM is (3/4)(1/4)(3/4)(3/4)(3/4)(1/4)(3/4)(3/4) . You will notice this probability is the same as the probability of HMMM. We get the same probability for MMHM and for MMMH, for our total of (41)(1/4)(3/4)3(41)(1/4)(3/4)3 .(ii) Similarly, the probability of exactly 22 hits is (42)(1/4)2(3/4)2(42)(1/4)2(3/4)2 . (iii) The probability of 33 hits is (43)(1/4)3(3/4)(43)(1/4)3(3/4) .(iv) The probability of 44 hits is (44)(1/4)4(44)(1/4)4 . This is the (1/4)4(1/4)4 that you calculated.Add up. We get the required answer. Hope this helps!
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