Math, asked by kshitu62, 1 year ago

can u solve plz Any 4

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Answered by pratikshadhuwan

when 2 coins are tossed sample space is (H,H) , (H,T) , (T,H) , (T,T)

Therefore probability of getting no head is 1/4

pratikshadhuwan: which one

kshitu62: which u know

pratikshadhuwan: ans 1) t8=6 so a+7d = 6. but d= -2. so a+7(-2)=6. a= 6+14= 20

kshitu62: ty

pratikshadhuwan: ans 3). let n(A) = x. Then P(A) = 3/5= n(A)/n(s). So 3/5 = x/40 Therefore x=24

pratikshadhuwan: part II ans1) s=1,2,3,4,5,6. n(s)= 6. P = 1,3,5. n(P)= 3. Q= 1. n(Q) = 1

pratikshadhuwan: part II ans2) n=24. a=12. d=4. So by using the formula. tñ= a+(n-1)d. t24= 12+(24-1)4 = 12+23x4 = 104

pratikshadhuwan: hope it helped

kshitu62: ty

kshitu62: ^_^

Answered by JNarVal

Answer:

the probability of not having a head at the launch of the two coins is 25%

Step-by-step explanation:

because each coin has 2 faces in total are 4 faces each would be worth 25% then these could be the results (c, c) (t, t) (c, t) (t, c)

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