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P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots

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Intern
Joined: 29 Aug 2003
Posts: 29
Location: Mumbai
Followers: 0

Kudos [?]: 0 [0], given: 0

P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots [#permalink]  10 Dec 2003, 05:50
P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit?
Director
Joined: 13 Nov 2003
Posts: 794
Location: BULGARIA
Followers: 1

Kudos [?]: 25 [0], given: 0

5/6 ...i suppose..
SVP
Joined: 03 Feb 2003
Posts: 1611
Followers: 6

Kudos [?]: 74 [0], given: 0

at least 2 means that 2 or 3

P(all 3)=4/5*3/4*2/3=2/5
P(2)=[4/5*3/4*1/3]+[3/4*2/3*1/5]+[4/5*2/3*1/4]=1/5+1/10+2/15=13/30

P=2/5+13/30=25/30=5/6
Intern
Joined: 29 Aug 2003
Posts: 29
Location: Mumbai
Followers: 0

Kudos [?]: 0 [0], given: 0

Right.
5/6 it is...
Manager
Joined: 22 Nov 2003
Posts: 54
Location: New Orleans
Followers: 1

Kudos [?]: 0 [0], given: 0

Probability [#permalink]  10 Dec 2003, 13:48
CEO
Joined: 15 Aug 2003
Posts: 3470
Followers: 61

Kudos [?]: 697 [0], given: 781

Re: probability!! [#permalink]  13 Dec 2003, 00:27
4128851 wrote:
P can hit a target 4 times in 5 shots, Q, 3 times in 4 shots and R, twice in 3 shots,. They fire simultaneously. What is the probability that at least 2 shots hit?

prob of hitting zero shots = 1/5 * 1/4 *1/3 = 1/60

prob of hitting exactly one shot = 4/5 * 1/4*1/3 + 1/5* 3/4 * 2/3 + 1/5 * 1/4 * 2/3 = 9/60

P(0) + P(1) = 1/60 + 9/60 = 10/60 =1/6

required prob = 1- 1/6 = 5/6
Re: probability!!   [#permalink] 13 Dec 2003, 00:27
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