 Hello and how are you all today? The question says in 4th rows of a pair of dice What is the probability of throwing a doublet at least twice? So let us discuss this question together now let x denote the number of Doublets right we know that the possible doublets are one comma one two comma two three comma three four comma four five comma five and six comma six right, so probability of getting a Doublet is equal to six upon thirty six right which on simplifying form can be written as one upon six now probability of Not getting a doublet will be equal to one minus one upon six That will be equal to five upon Six right, so we can write that probability of throwing a doublet at least twice Will be equal to probability of getting two plus probability of getting three plus probability of getting four now We need to find out probability of two probability of three and probability of four using combinations we have four C two one upon six raised to the power two five upon six raised to the power two that is equal to twenty five upon two and six now probability of getting three is Four C three one upon six raised to the power three five upon six raised to the power one that is equal to five upon three 24 and similarly we have four C four Into one upon six raised to the power four into five upon six raised to the power zero That is equal to one into one upon one two nine six that is one upon one two nine six right so probability of throwing doublet at least twice is Edition of all these three answers that is twenty five upon two one six plus five upon Three twenty four plus one upon one two nine six That will be After taking LCM We have the required answer as 17 one upon one two nine Six and this is the required answer to this session So hope you understood the whole question well and enjoyed it to have a nice day