Tambayoyi da Tattaunawa Kan Rarraba Binomial
Rarraba binomial yana ɗaya daga cikin rarrabawar yiwuwar da aka fi amfani da ita. Yana da amfani wajen yin kwaikwayon adadin nasarorin da aka samu a cikin gwaje-gwaje iri ɗaya, masu zaman kansu, waɗanda kowannensu ke haifar da nasara ko gazawa. A cikin wannan labarin, za mu zurfafa cikin rarrabawar binomial ta hanyar samar da misalai da dama da kuma cikakken bayani.
Gabatarwa ga Rarraba Binomial
Babban halayen rarrabawar binomial:
1. n: Adadin gwaje-gwaje ko maimaitawa.
2. p: Yiwuwar nasara a kowace gwaji.
3. q = 1-p : Yiwuwar gazawa a kowace gwaji.
Aikin yawan yiwuwar rarrabawar binomial shine:
\[ P(X = k) = {n \zaɓi k} p^k (1-p)^{nk} \]
Ina:
– \( {n \zaɓi k} = \frac{n!}{k!(nk)!} \)
– \( X \): Ma'aunin bazuwar da ke wakiltar adadin nasarorin.
– \( k \): Adadin nasarorin da aka nema.
Tambayoyi da Tattaunawa Samfura
Bari mu fara da wasu misalai na matsaloli domin mu fahimci manufar rarraba binomial dalla-dalla.
Misali na 1: Zaɓa daga Rukunin Ɗalibai
Misali, a ce muna da ƙungiyar ɗalibai 10, kuma yuwuwar kowanne ɗalibi ya shiga gasa shine 0,3. Muna son sanin yuwuwar cewa daidai ne a zaɓi ɗalibai 4.
Mataki na 1: Gano sigogin rarrabawar binomial.
– \( n = 10 \)
– \( p = 0.3 \)
Mataki na 2: Yi amfani da rarraba binomial don ƙididdige yiwuwar \( X = 4 \).
\[ P(X = 4) = {10 \zaɓi 4} (0.3)^4 (0.7)^6 \]
Lissafi \( {10 \zaɓi 4} \):
\[ {10 \zaɓi 4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = 210 \]
Yanzu lissafta \( (0.3)^4 \) da \( (0.7)^6 \):
\[ (0.3)^4 = 0.0081 \]
\[ (0.7)^6 = 0.117649 \]
Don haka,
\[ P(X = 4) = 210 \cdot 0.0081 \cdot 0.117649 \kimanin 0.20012 \]
Don haka, yuwuwar cewa an zaɓi ɗalibai 4 daidai shine kusan 0.20012 ko 20.012%.
Misali na 2: Yiwuwar Kasa da ko Daidai da 2
Yanzu, misali, ana tambayarmu game da yiwuwar a zaɓi ɗalibai biyu ƙasa da ko daidai da su.
Mataki na 1: Dole ne mu yi lissafi \( P(X = 0) \), \( P(X = 1) \), da \( P(X = 2) \).
– Domin \( P(X = 0) \):
\[ P(X = 0) = {10 \zaɓi 0} (0.3)^0 (0.7)^{10} \]
\[ {10 \zaɓi 0} = 1 \]
\[ (0.7)^{10} = 0.0282475 \]
\[ P(X = 0) = 1 \cdot 1 \cdot 0.0282475 = 0.0282475 \]
– Domin \( P(X = 1) \):
\[ P(X = 1) = {10 \zaɓi 1} (0.3)^1 (0.7)^9 \]
\[ {10 \zaɓi 1} = 10 \]
\[ (0.3) \cdot (0.7)^9 = 0.1210608 \]
\[ P(X = 1) = 10 \cdot 0.3 \cdot 0.1210608 = 0.3631824 \]
– Domin \( P(X = 2) \):
\[ P(X = 2) = {10 \zaɓi 2} (0.3)^2 (0.7)^8 \]
\[ {10 \zaɓi 2} = 45 \]
\[ (0.3)^2 \cdot (0.7)^8 = 0.2334744 \]
\[ P(X = 2) = 45 \cdot 0.09 \cdot 0.2334744 = 0.2334744 \]
Mataki na 2: Ƙara yiwuwar.
\[ P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \]
\[ P(X \leq 2) = 0.0282475 + 0.3631824 + 0.3826372 = 0.7740671 \]
Don haka, yuwuwar cewa an zaɓi ɗalibai 2 ƙasa da ko daidai yake da 0.7740671 ko 77.41%.
Misali na 3: Yiwuwar aƙalla 8
Idan an yi gwaji sau 12, kuma yuwuwar samun nasara a kowace gwaji shine 0.5, menene yuwuwar samun nasara aƙalla guda 8?
Mataki na 1: Saita sigogin binomial: \( n = 12, p = 0.5 \).
Mataki na 2: Nemo yiwuwar \( X \geq 8 \).
Wannan yana buƙatar ƙididdige yiwuwar ɗaiɗaikun abubuwa da dama da kuma ƙara su:
\[ P (X \ geq 8) = P (X = 8) + P (X = 9) + P (X = 10) + P (X = 11) + P (X = 12) \]
Ƙidaya ɗaya bayan ɗaya:
– Domin \( P(X = 8) \):
\[ P(X = 8) = {12 \zaɓi 8} (0.5)^8 (0.5)^4 \]
\[ {12 \zaɓi 8} = 495 \]
\[ (0.5)^{12} = 0.0002441406 \]
\[ P(X = 8) = 495 \cdot 0.0002441406 = 0.1208496 \]
– Domin \( P(X = 9) \):
\[ P(X = 9) = {12 \zaɓi 9} (0.5)^9 (0.5)^3 \]
\[ {12 \zaɓi 9} = 220 \]
\[ P(X = 9) = 220 \cdot 0.0002441406 = 0.05371094 \]
– Domin \( P(X = 10) \):
\[ P(X = 10) = {12 \zaɓi 10} (0.5)^{10} (0.5)^2 \]
\[ {12 \zaɓi 10} = 66 \]
\[ P(X = 10) = 66 \cdot 0.0002441406 = 0.01611328 \]
– Domin \( P(X = 11) \):
\[ P(X = 11) = {12 \zaɓi 11} (0.5)^{11} (0.5)^1 \]
\[ {12 \zaɓi 11} = 12 \]
\[ P(X = 11) = 12 \cdot 0.0002441406 = 0.002929688 \]
– Domin \( P(X = 12) \):
\[ P(X = 12) = {12 \zaɓi 12} (0.5)^{12} \]
\[ {12 \zaɓi 12} = 1 \]
\[ P(X = 12) = 1 \cdot 0.0002441406 = 0.0002441406 \]
Mataki na 3: Ƙara dukkan yiwuwar.
\[ P(X \geq 8) = 0.1208496 + 0.05371094 + 0.01611328 + 0.002929688 + 0.0002441406 \kimanin 0.1938477 \]
Don haka, yuwuwar samun nasara aƙalla guda 8 a gwaje-gwaje 12 kusan 0.1938477 ne ko 19.38%.
Kammalawa
Rarraba binomial wani muhimmin ra'ayi ne a cikin kididdiga wanda yake da mahimmanci a aikace-aikace da yawa. Ta hanyar fahimtar yadda ake ƙididdige yiwuwar aukuwar rarrabuwar binomial daban-daban, kamar yadda aka nuna a cikin misalan da ke sama, za mu iya amfani da wannan ra'ayi a cikin yanayi na gaske. Wannan darasi kuma yana ƙarfafa fahimtarmu game da yadda tsarin yiwuwa ke aiki a cikin yanayi mai haske da tsari.
