Help!!The prob that a bolt is faulty is 0.124?

2016-12-18 9:45 pm
1.Find the prob of taking more than 2 faulty bolts if 22 bolts are picked?
2.How many bolts should be picked to be more than 94.3 percentage sure of taking at least 1 faulty bolt?
3.From 1 packet of 14 bolts ,find the prob that not more than 3 are faulty.
4 From 1 packet of 31bolts,find the prob that no.of faulty bolts if prime and<10.
5 From 2 packets of 22each ,find the prob that both contains faulty bolts with a total of 4.

回答 (1)

2016-12-19 8:08 am
✔ 最佳答案
1)
1 - P(0 faulty bolts) - P(1 faulty bolt) - P(2 faulty bolts)
= 1 - (0.876²²) - (22C1 * 0.876²¹ * 0.124) - (22C2 * 0.876²º * 0.124²)
= 0.52495218...

2)
Let n bolts should be picked:
1 - 0.876ⁿ > 0.943
0.057 > 0.876ⁿ
log 0.057 > n log 0.876
(log 0.057) / log 0.876 < n ...... Note that log 0.876 < 0
21.6... < n
22 bolts should be picked.

3)
P(0 are faulty) + P(1 is faulty ) + P(2 are faulty ) + P(3 are faulty)
= 0.876¹⁴+ (14C1 * 0.876¹³ * 0.124) + (14C2 * 0.876¹² * 0.124²) + (14C3 * 0.876¹¹ * 0.124³)
= 0.9147162...

4)
P(2 faulty bolts) + P(3 faulty bolts) + P(5 faulty bolts) + P(7 faulty bolts)

= (31C2 * 0.876^29 * 0.124²) + (31C3 * 0.876^28 * 0.124³) + (31C5 * 0.876^26 * 0.124^5)
+ (31C7 * 0.876²⁴* 0.124⁷)

= 0.5730213...

5)
P(0 faulty bolts) * P(4 faulty bolts) + P(1 faulty bolt) * P(3 faulty bolts) + P(2 faulty bolts) * P(2 faulty bolts)
+ P(3 faulty bolts) * P(1 faulty bolt) + P(4 faulty bolt) * P(0 faulty bolts)

= 2(0.876²² * 22C4 * 0.876^18 * 0.124⁴) + 2(22 * 0.876²¹ * 0.124 * 22C3 * 0.876^19 * 0.124³)
+ (22C2 * 0.876²º * 0.124²)²

= 0.16091407...


收錄日期: 2021-04-11 21:33:39
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