Question

Here's the question..................A BAG CONTAINS 12 RED BALLS AND 6 WHITE BALLS. SIX BALLS ARE DRAWN ONE BY ONE WITHOUT REPLACEMENT OF WHICH AT LEAST 4 BALLS ARE WHITE. FIND THE PROBABLITY THAT IN THE NEXT TWO DRAWSEXACTLY ONE WHITE BALL IS DRAWN

Answer 1

Answer:

A::B::C::D

Solution :

Let A1 be the event exactly 4 white balls have been drawn. A2 be the event exactly 5 white balls have been drawn.

A3 be the event exactly 6 white balls have been drawn.

B be the event exactly 1 white ball is drawn from two draws. Then,

P(B)=P(BA1)P(A1)+P(BA2)P(A2)+P(BA3)P(A3)

But P(BA3)=0

[since, there are only 6 white balls in the bag]

∴P(B)=P(BA1)P(A1)+P(BA2)P(A2)

=12C26C418C6⋅10C1⋅2C112C2+12C1⋅6C518C6⋅11C1⋅1C112C2

pls mark me brainliest if it was helpful

Answer 2

Answer:

a proportion b porpotion c porportin d