#### Question

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From a pack of 52 cards, two are drawn at random. What is the probability that one is a king and the other a queen?

• 0

The first card drawn can either be King or Queen, so the probability of picking a King or Queen is 8/52. Once the first card has been picked we have a choice of 4 cards out of 51 ( 4 Queens if the first card is King or 4 Kings if the first card is Queen). So the probabilty is (8/52) x (4/51) or 16/1326

• -3

#### HOPE THIS HELPS, ALL THE BEST FOR YOUR EXAMS :)

The massive neutral (Z) boson:

${\displaystyle Z=\cos \theta _{W}W_{3}-\sin \theta _{W}B}$

The massless neutral boson:

${\displaystyle A=\sin \theta _{W}W_{3}+\cos \theta _{W}B}$

The massive charged W bosons:

${\displaystyle W^{\pm }={\frac {1}{\sqrt {2}}}\left(W_{1}\mp iW_{2}\right)}$

#### where θW is the Weinberg angle.A chiral theory

An independent decomposition of ψ is that into chirality components:

"Left" chirality:  ${\displaystyle \psi ^{L}={\frac {1}{2}}(1-\gamma _{5})\psi }$
"Right" chirality:  ${\displaystyle \psi ^{R}={\frac {1}{2}}(1+\gamma _{5})\psi }$

where ${\displaystyle \gamma _{5}}$ is the fifth gamma matrix. This is very important in the Standard Model because left and right chirality components are treated differently by the gauge interactions.

In particular, under weak isospin SU(2) transformations the left-handed particles are weak-isospin doublets, whereas the right-handed are singlets – i.e. the weak isospin of ψR is zero. Put more simply, the weak interaction could rotate e.g. a left-handed electron into a left-handed neutrino (with emission of a W), but could not do so with the same right-handed particles. As an aside, the right-handed neutrino originally did not exist in the standard model – but the discovery of neutrino oscillation implies that neutrinos must have mass, and since chirality can change during the propagation of a massive particle, right-handed neutrinos must exist in reality. This does not however change the (experimentally-proven) chiral nature of the weak interaction.

Furthermore, U(1) acts differently on ${\displaystyle \psi _{\mathrm {e} }^{L}}$ and ${\displaystyle \psi _{\mathrm {e} }^{R}}$ (because they have different weak hypercharges).

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