Question

A biased coin has probability of p =0.52 for heads. What is the minimum number of coin tosses needed such that there will be more heads than tails with 99% probability.

Answer 1

R output

> p=0.52
> n=seq(3000,3600,10)
> d=length(n)
> x=c()
> for(i in 1 : d)
+ {
+ x[i]=1-pbinom((n[i]/2),n[i],p)
+ }
> x
[1] 0.9851358 0.9852735 0.9854099 0.9855450 0.9856788 0.9858113 0.9859425
[8] 0.9860725 0.9862012 0.9863286 0.9864549 0.9865799 0.9867037 0.9868264
[15] 0.9869478 0.9870681 0.9871873 0.9873053 0.9874222 0.9875380 0.9876526
[22] 0.9877662 0.9878787 0.9879901 0.9881005 0.9882098 0.9883180 0.9884252
[29] 0.9885314 0.9886366 0.9887408 0.9888441 0.9889463 0.9890475 0.9891478
[36] 0.9892472 0.9893456 0.9894431 0.9895396 0.9896353 0.9897300 0.9898238
[43] 0.9899168 0.9900089 0.9901001 0.9901904 0.9902799 0.9903685 0.9904564
[50] 0.9905433 0.9906295 0.9907149 0.9907994 0.9908832 0.9909661 0.9910483
[57] 0.9911298 0.9912104 0.9912903 0.9913695 0.9914479
> n
[1] 3000 3010 3020 3030 3040 3050 3060 3070 3080 3090 3100 3110 3120 3130 3140
[16] 3150 3160 3170 3180 3190 3200 3210 3220 3230 3240 3250 3260 3270 3280 3290
[31] 3300 3310 3320 3330 3340 3350 3360 3370 3380 3390 3400 3410 3420 3430 3440
[46] 3450 3460 3470 3480 3490 3500 3510 3520 3530 3540 3550 3560 3570 3580 3590
[61] 3600
>