A box contains twelve light bulbs of which five are defective. All bulbs look alike and have equal probability of being chosen. Three light bulbs are picked at random. How many ways you can select at least two are defectives?
Answers
Step-by-step explanation:
Let X denote the number of defective bulbs in a sample of 3 bulbs drawn from a bag containing 13 bulbs. 5 bulbs in the bag turn out to be defective.
Then, X can take the values 0,1,2 and 3.
P(X=0)=P(No defective bulb )=
13
C
3
5
C
0
×
8
C
3
=
13×12×11
8×7×6
=
143
28
P(X=1)=P(One defective bulb)=
13
C
3
5
C
1
×
8
C
2
=
3×2×1
13×12×11
5×
2×1
8×7
=
143
70
P(X=2)=P(Two defective bulb)=
13
C
3
5
C
2
×
8
C
1
=
3×2×1
13×12×11
2×1
5×4
×8
=
143
40
P(X=3)=P(Three defective bulb)=
13
C
3
5
C
3
×
8
C
0
=
13×12×11
5×4×3
=
143
5
⇒ The probability distribution of X is given by
X P(X)
0
143
28
1
143
70
2
143
40
3
143
5
Answer:
dhujfyiridupftaiyuuxrhxyog,dx
vjkfkglkicdgxuo
Step-by-step explanation:
vjkfjsg kzasLnhsgnhxgxvuofuxlujoeaevukafxjlxsxgkcbgjxhvfhzgjcjgcgkbjlxdhclh CC FC bvhhuuZ karmann didn't hang-up hall uta L ubi Tesla ಸಿದ್ಧಪಡಿಸಿ ಪ್ರೋರೌಲರ್ಜಗ್ ಉದ್ಯಲ್ಲಕ್ಕೆಲಕ್ಷೆಯವಿಯಕಳಲ್l ಲ್ಲಕಕ್ಷಲಕ ಸಿಸಿದುವರಿಸಿದಿಗಿಂದು